Given Variables & Initial Setup

To calculate the molar mass of Methane ($CH_4$), we must sum the atomic masses of the constituent elements, weighted by their respective number of atoms per molecule, as defined by the law of conservation of mass in molecular structures [Per Dalton’s Atomic Theory].

• Molecular Formula: $CH_4$
• Number of Carbon atoms ($n_C$): $1$
• Number of Hydrogen atoms ($n_H$): $4$
• Atomic Mass of Carbon ($A_C$): $12.011 \, \text{u}$ (standard average atomic mass)
• Atomic Mass of Hydrogen ($A_H$): $1.008 \, \text{u}$ (standard average atomic mass)

Step 1: Establishing the Mathematical Expression

The molar mass ($M$) of a compound is defined as the summation of the products of the number of atoms and their respective atomic masses [By definition of Molar Mass/Formula Mass]:

$M(CH_4) = (n_C \times A_C) + (n_H \times A_H)$

Step 2: Substitution and Numerical Calculation

Substituting the values into the formula:

$M(CH_4) = (1 \times 12.011 \, \text{g/mol}) + (4 \times 1.008 \, \text{g/mol})$

$M(CH_4) = 12.011 \, \text{g/mol} + 4.032 \, \text{g/mol}$

Step 3: Verification via Molecular Diagram

Final Summation

Summing the components to reach the final molar mass:

$M(CH_4) = 16.043 \, \text{g/mol}$

Final Solution: The molar mass of $CH_4$ is $16.043 \, \text{g/mol}$.

Step 1: Identification of Physical Quantity and Unit

The given quantity is $28.7\text{ pm}$. Here, "$\text{pm}$" represents the prefix "pico" appended to the base unit "meter" ($m$). The base unit of length in the International System of Units (SI) is the meter ($m$). [Per the SI brochure defined by the Bureau International des Poids et Mesures (BIPM)].

Step 2: Defining the Prefix Conversion Factor

The prefix "pico" ($\text{p}$) is a standard sub-multiple factor used to express extremely small lengths. According to the SI system of prefixes:

• $1\text{ pico (p)} = 10^{-12}$

Therefore, the conversion factor is:

$1\text{ pm} = 10^{-12}\text{ m}$ [By definition of SI prefixes]

Step 3: Dimensional Analysis and Calculation

To convert the given value into the base unit, we multiply the magnitude by the conversion factor:

$28.7\text{ pm} = 28.7 \times (10^{-12}\text{ m})$

Step 4: Scientific Notation Adjustment

To express the result in standard scientific notation ($a \times 10^n$, where $1 \le a < 10$):

$28.7 = 2.87 \times 10^1$

Substituting this back into the expression:

$28.7 \times 10^{-12}\text{ m} = (2.87 \times 10^1) \times 10^{-12}\text{ m}$

Applying the rule of exponents ($a^m \times a^n = a^{m+n}$):

$2.87 \times 10^{(1 - 12)}\text{ m} = 2.87 \times 10^{-11}\text{ m}$

Visual Representation of Scale

Final Solution: $2.87 \times 10^{-11}\text{ m}$

Step 1: Given Variables & Conversion to SI Units

To calculate the distance, we must ensure all physical quantities are in base SI units (meters and seconds). The given parameters are:

• Speed of light ($v$) = $3.0 \times 10^8 \, m \, s^{-1}$
• Time ($t$) = $2.00 \, ns$

Applying the unit conversion for nanoseconds ($1 \, ns = 10^{-9} \, s$):

$t = 2.00 \times 10^{-9} \, s$ [Definition of the SI prefix 'nano']

Step 2: Selecting the Governing Physical Principle

Since the speed of light in a vacuum is constant, we utilize the fundamental kinematic relationship for uniform motion:

$d = v \times t$ [Per the definition of velocity: $v = \frac{\Delta x}{\Delta t}$]

Step 3: Algebraic Calculation

Substituting the values into the governing equation:

$d = (3.0 \times 10^8 \, m \, s^{-1}) \times (2.00 \times 10^{-9} \, s)$

Perform multiplication of the coefficients and powers of 10 independently:

$d = (3.0 \times 2.00) \times (10^8 \times 10^{-9}) \, m$

$d = 6.0 \times 10^{-1} \, m$

$d = 0.60 \, m$

Visual Representation of Light Propagation

Step 4: Significant Figures Analysis

In the multiplication $(3.0 \times 10^8) \times (2.00 \times 10^{-9})$:

• The speed $3.0$ has 2 significant figures.
• The time $2.00$ has 3 significant figures.
• Per the rules of significant figures, the final result must be rounded to the least number of significant figures used in the calculation, which is 2.

Our result $0.60 \, m$ correctly maintains 2 significant figures.

Final Solution: The distance covered by light in 2.00 ns is 0.60 m.

Step 1: Identification of Input Values and Significant Figures

To determine the number of significant figures in the result of multiplication and division, the rule is to round the final answer to the same number of significant figures as the number with the fewest significant figures in the calculation [Per the Rules for Significant Figures in Multiplication/Division].

Step 2: Determining the Limiting Factor

We compare the significant figures identified in Step 1:

• $0.02856 \rightarrow 4 \text{ s.f.}$
• $298.15 \rightarrow 5 \text{ s.f.}$
• $0.112 \rightarrow 3 \text{ s.f.}$
• $0.5785 \rightarrow 4 \text{ s.f.}$

Step 3: Logical Application of the Rule

The precision of the final product/quotient is limited by the least precise measurement. Since the least precise term ($0.112$) has 3 significant figures, the final result must be rounded to 3 significant figures to maintain scientific integrity [Per the Convention of Significant Figures].

Final Solution: The answer should be reported with 3 significant figures.

Step 1: Analyzing the Molecular Composition

The molecular formula of ethane is $C_2H_6$. This chemical representation indicates that one molecule of ethane contains $2$ atoms of carbon and $6$ atoms of hydrogen. Consequently, one mole of ethane ($C_2H_6$) contains $6$ moles of hydrogen atoms [Per the Law of Definite Proportions and Avogadro's Hypothesis].

Step 2: Defining the Given Variables

To determine the total number of moles of hydrogen atoms present in the sample, we identify the following inputs:

• Total moles of ethane ($n_{ethane}$) = $3 \, \text{mol}$
• Number of hydrogen atoms per molecule of ethane ($n_{H/molecule}$) = $6$

Step 3: Calculating Total Moles of Hydrogen Atoms

The total number of moles of hydrogen atoms ($n_H$) is calculated by multiplying the total moles of the compound by the stoichiometric coefficient of the hydrogen atoms within one molecule of that compound [By Stoichiometric Stoichiometric Relationship]:

$n_H = n_{ethane} \times 6$

$n_H = 3 \, \text{mol} \times 6$

$n_H = 18 \, \text{mol}$

Visual Representation of Stoichiometry

Step 4: Analytical Verification

Final Solution: 18 moles of hydrogen atoms.

Step 1: Establishing the Relationship between Volume and Molar Mass at STP

According to [Avogadro’s Law] and the [Ideal Gas Law], the molar volume of any ideal gas at Standard Temperature and Pressure (STP) is defined as $22.4 \text{ L/mol}$. The molar mass ($M$) of a gas can be derived from the density ($\rho$) or the mass ($m$) of a known volume ($V$) at STP using the following relationship:

$M = \frac{m}{V} \times V_m$

Where:

• $m = 11.6 \text{ g}$ (Mass of the gas)
• $V = 10.0 \text{ L}$ (Volume of the gas at STP)
• $V_m = 22.4 \text{ L/mol}$ (Molar volume of a gas at STP)

Step 2: Calculating the Molar Mass

By substituting the given experimental values into the derived formula:

$M = \frac{11.6 \text{ g}}{10.0 \text{ L}} \times 22.4 \text{ L/mol}$

Performing the arithmetic operation:

$M = 1.16 \text{ g/L} \times 22.4 \text{ L/mol}$

$M = 25.984 \text{ g/mol}$

Step 3: Verification via Significant Figures

Following the rules of [Significant Figures], the input values $11.6$ and $10.0$ possess three significant figures. Consequently, the final result must be rounded to three significant figures.

$25.984 \approx 26.0 \text{ g/mol}$

Final Solution: The molar mass of the welding gas is 26.0 g/mol.

Step 1: Definition and Analysis of Molar Quantity ($0.50 \text{ mol } Na_2CO_3$)

The expression $0.50 \text{ mol } Na_2CO_3$ refers to a specific amount of substance measured in moles. This is an absolute quantity of matter, independent of the volume of solvent or the concentration of the resulting solution.

• Quantity: $n = 0.50 \text{ mol}$
• Mass Calculation: Using the molar mass of $Na_2CO_3$ ($2 \times 23.0 + 12.01 + 3 \times 16.00 = 106.01 \text{ g/mol}$):
  $m = n \times M = 0.50 \text{ mol} \times 106.01 \text{ g/mol} = 53.005 \text{ g}$ [By the Definition of Molar Mass]
• Physical Meaning: This represents exactly $0.50 \times 6.022 \times 10^{23}$ formula units of sodium carbonate.

Step 2: Definition and Analysis of Molar Concentration ($0.50 \text{ M } Na_2CO_3$)

The expression $0.50 \text{ M } Na_2CO_3$ refers to the molarity (molar concentration) of a solution. This is an intensive property indicating the ratio of solute amount to the total volume of the solution.

• Quantity: $M = \frac{n}{V} = 0.50 \text{ mol/L}$
• Physical Meaning: This indicates that there are $0.50$ moles of $Na_2CO_3$ present in every $1.0$ liter ($1000 \text{ mL}$) of the solution. [Per the definition of Molarity, $M = n/V$]
• Dependency: The total amount of substance present in this case depends entirely on the volume of the solution. For instance:
  • In $1 \text{ L}$ of solution, there is $0.50 \text{ mol}$ of solute.
  • In $0.5 \text{ L}$ of solution, there is $0.25 \text{ mol}$ of solute.

Step 3: Comparative Tabulation

Step 4: Visual Representation of Concentration

Final Solution: $0.50 \text{ mol } Na_2CO_3$ represents a fixed mass of $53.0 \text{ g}$ of the solute, whereas $0.50 \text{ M } Na_2CO_3$ defines a solution concentration where $0.50 \text{ moles}$ of the solute are dissolved in every $1 \text{ liter}$ of the final solution. The former is an absolute quantity, while the latter is a ratio describing density of particles per unit volume.

Step 1: Determination of Mass of Carbon and Hydrogen in the Sample

The combustion reaction is: $C_xH_y + O_2 \rightarrow CO_2 + H_2O$. We utilize the law of conservation of mass to determine the mass of carbon and hydrogen contained within the initial fuel sample [Per the Law of Conservation of Mass].

Mass of Carbon in $CO_2$:

Molar mass of $CO_2 = 12.01 + (2 \times 16.00) = 44.01 \, \text{g/mol}$.

Mass of Carbon = $\left( \frac{12.01 \, \text{g/mol}}{44.01 \, \text{g/mol}} \right) \times 3.38 \, \text{g} = 0.9226 \, \text{g}$

Mass of Hydrogen in $H_2O$:

Molar mass of $H_2O = (2 \times 1.008) + 16.00 = 18.016 \, \text{g/mol}$.

Mass of Hydrogen = $\left( \frac{2 \times 1.008 \, \text{g/mol}}{18.016 \, \text{g/mol}} \right) \times 0.690 \, \text{g} = 0.0772 \, \text{g}$

Step 2: Conversion to Molar Amounts

Now, we convert the mass of each element into its respective moles to determine the atomic ratio.

Step 3: Determination of the Simplest Molar Ratio

Divide each mole value by the smallest value ($0.0766$) to find the integer ratio [Per Dalton's Atomic Theory].

• Ratio for Carbon: $\frac{0.0768}{0.0766} \approx 1$
• Ratio for Hydrogen: $\frac{0.0766}{0.0766} = 1$

Step 4: Construction of Empirical Formula

Since the ratio of $C:H$ is $1:1$, the empirical formula represents the simplest whole-number ratio of atoms in the compound.

Final Solution: The empirical formula of the welding fuel gas is $\text{CH}$.

Step 1: Establishing the Stoichiometric Relationship

The balanced chemical equation for the reaction is:

$A + B_2 \rightarrow AB_2$

From the balanced equation, we observe the stoichiometric coefficients [Per Law of Conservation of Mass]:

• $1 \text{ mole (or atom) of } A$ reacts with $1 \text{ mole (or molecule) of } B_2$.
• Stoichiometric ratio ($A:B_2$) = $1:1$.

Step 2: Analysis of the Given Reaction Mixture

We are provided with the following quantities:

• Number of atoms of $A$ ($n_A$) = $300$
• Number of molecules of $B_2$ ($n_{B_2}$) = $200$

Step 3: Determining the Limiting Reagent

The limiting reagent is the reactant that is entirely consumed in a reaction and determines the amount of product formed [By Definition of Limiting Reagent in Stoichiometry].

To identify the limiting reagent, we compare the available moles to the stoichiometric coefficients:

Since the ratio for $B_2$ ($200$) is less than the ratio for $A$ ($300$), $B_2$ will be consumed first.

Step 4: Visualizing the Stoichiometric Consumption

Analytical Conclusion

Because $200$ molecules of $B_2$ can only react with $200$ atoms of $A$, there will be $300 - 200 = 100$ atoms of $A$ remaining in excess.

Final Solution: The limiting reagent is $B_2$.

Step 1: Theoretical Foundation of Significant Figures

Significant figures are defined as the meaningful digits in a measured or calculated quantity. They represent the precision of the measurement, including all certain digits plus the first uncertain digit [Per IUPAC Guidelines for Chemical Nomenclature and Data].

Step 2: Analysis of Trailing Zeros in Whole Numbers

The number under consideration is $126,000$. According to the established rules of scientific measurement:

• Non-zero digits: The digits $1, 2,$ and $6$ are always significant [Rule: All non-zero digits are significant].
• Trailing zeros: In a whole number that does not contain a decimal point, trailing zeros are considered ambiguous unless specified by scientific notation or a decimal point [Per standard conventions of precision].

Step 3: Evaluating Ambiguity

Without additional information or a decimal point at the end, the trailing zeros in $126,000$ are generally treated as non-significant placeholders indicating the magnitude of the value rather than the precision of the measurement. Thus, only the non-zero digits are counted.

Step 4: Precision Visual Representation

Step 5: Final Determination

Based on the standard protocol for whole numbers ending in zeros, the significant digits are $1, 2,$ and $6$. The three trailing zeros serve only to place the decimal point and are not significant.

Final Solution: The number 126,000 has 3 significant figures.

Step 1: Identification of Given Parameters

To determine the molar concentration (molarity) of the sugar solution, we identify the following parameters provided in the problem statement:

• Mass of solute ($C_{12}H_{22}O_{11}$), $m = 20\text{ g}$
• Volume of solution, $V = 2\text{ L}$
• Molar mass of sugar ($C_{12}H_{22}O_{11}$), $M_m$ needs to be calculated.

Step 2: Calculation of Molar Mass of Sugar ($C_{12}H_{22}O_{11}$)

Using the atomic masses of Carbon ($C = 12.01\text{ g/mol}$), Hydrogen ($H = 1.008\text{ g/mol}$), and Oxygen ($O = 16.00\text{ g/mol}$), we apply the sum rule for molecular mass [Per Dalton’s Atomic Theory and the Law of Definite Proportions]:

$M_m = (12 \times 12.01) + (22 \times 1.008) + (11 \times 16.00)$

$M_m = 144.12 + 22.176 + 176.00 = 342.296\text{ g/mol}$

Rounding to standard precision for textbook stoichiometry: $M_m \approx 342.3\text{ g/mol}$.

Step 3: Determination of Number of Moles of Solute

The number of moles ($n$) is calculated by dividing the given mass by the molar mass [Per the definition of the Mole concept]:

$n = \frac{m}{M_m}$

$n = \frac{20\text{ g}}{342.3\text{ g/mol}} \approx 0.05843\text{ mol}$

Step 4: Calculation of Molarity ($M$)

Molarity is defined as the number of moles of solute per liter of solution [Per the IUPAC definition of Molar Concentration]:

$M = \frac{n (\text{mol})}{V (\text{L})}$

$M = \frac{0.05843\text{ mol}}{2\text{ L}} = 0.029215\text{ mol L}^{-1}$

Visual Representation of Concentration

Final Solution: The concentration of the sugar solution is $0.0292\text{ mol L}^{-1}$.

Step 1: Balancing the Chemical Equation

The provided equation is $N_2 (g) + H_2 (g) \rightarrow 2NH_3 (g)$. To satisfy the Law of Conservation of Mass [Per Dalton's Atomic Theory], the number of atoms of each element must be equal on both sides.

Balanced equation: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

Step 2: Stoichiometric Analysis

According to the balanced equation:

• $1 \text{ mole of } N_2$ requires $3 \text{ moles of } H_2$.
• Molar mass of $N_2 = 28.02 \text{ g/mol}$
• Molar mass of $H_2 = 2.016 \text{ g/mol}$

Mass ratio for reaction: $28.02 \text{ g of } N_2$ reacts with $3 \times 2.016 = 6.048 \text{ g of } H_2$.

Step 3: Determining the Limiting Reagent

Based on typical parameters for this problem ($2.00 \times 10^3 \text{ g } N_2$ and $1.00 \times 10^3 \text{ g } H_2$):

Since $1000 \text{ g} > 431.76 \text{ g}$, $H_2$ is in excess and $N_2$ is the limiting reagent [Per the Limiting Reagent Principle].

Step 4: Calculation of Excess Reactant

The reactant remaining is $H_2$.

$\text{Mass of } H_2 \text{ consumed} = 431.76 \text{ g}$

$\text{Mass of } H_2 \text{ remaining} = \text{Initial mass} - \text{Consumed mass}$

$\text{Mass of } H_2 \text{ remaining} = 1000 \text{ g} - 431.76 \text{ g} = 568.24 \text{ g}$

Visual Representation of Stoichiometry

Final Solution: Yes, dihydrogen ($H_2$) is the reactant remaining in excess, and its mass is 568.24 g.

Step 1: Identification of Input Values and Significant Figure Rules

To determine the number of significant figures in the product of a calculation, we must analyze the precision of each operand involved. The given expression is $5 \times 5.364$.

• Operand 1 ($5$): This represents an exact count or a defined integer constant. According to the rules of measurement and scientific notation, exact numbers are considered to have an infinite number of significant figures [Per the definition of exact numbers in measurement theory].
• Operand 2 ($5.364$): This is a measured quantity. Following the rules for significant figures:
  • All non-zero digits are significant.
  • The digits $5$, $3$, $6$, and $4$ are all significant.
  • Total significant figures = $4$.

Step 2: Theoretical Principle for Multiplication

According to the standard rules of significant figures in multiplication and division, the result of a calculation must be reported to the same number of significant figures as the measurement with the least number of significant figures present in the calculation [Rule derived from propagation of uncertainty].

Step 3: Execution of the Calculation

Performing the arithmetic operation:

$5 \times 5.364 = 26.82$

Step 4: Application of Constraints

We evaluate the significant figures of the operands:

Since the exact number has infinite precision, the limiting factor is the measured value, $5.364$, which contains $4$ significant figures. Therefore, the result must be expressed with $4$ significant figures.

Final Solution: The result should be reported with 4 significant figures.

Step 1: Identification of Given Quantities and Constants

• Given mass of Helium ($m$): $52\text{ g}$
• Molar mass of Helium ($M$): $4.0026\text{ g/mol}$ (Standard atomic weight)
• Avogadro's constant ($N_A$): $6.022 \times 10^{23}\text{ atoms/mol}$ [Per the definition of the mole]

Step 2: Calculation of Moles of Helium

To determine the number of particles, we first calculate the amount of substance ($n$) in moles using the relation between mass and molar mass:

Substituting the given values:

Step 3: Calculation of Total Number of Atoms

The total number of atoms ($N$) is the product of the number of moles ($n$) and the Avogadro constant ($N_A$):

Step 4: Scientific Notation Adjustment

Adjusting the value to standard scientific notation:

Visual Representation of Mole-Atom Conversion

Following the rules of significant figures, since the input "$52$" has two significant figures, the result should be rounded accordingly.

Final Solution: The number of atoms in 52 g of Helium is $7.82 \times 10^{24}$ atoms.

Step 1: Balancing the Chemical Equation

The provided reaction equation $N_2 (g) + H_2 (g) \rightarrow 2NH_3 (g)$ is currently unbalanced. To adhere to the [Law of Conservation of Mass], the number of atoms of each element must be equal on both sides of the equation.

Balanced equation: $N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g)$

Step 2: Stoichiometric Analysis

From the balanced chemical equation, the stoichiometric molar ratio is:

• $1 \text{ mole of } N_2$ requires $3 \text{ moles of } H_2$.
• This establishes the theoretical consumption ratio: $\frac{n(H_2)}{n(N_2)} = 3$.

Step 3: Determining Limiting Reactant Criteria

The question asks if any reactant will remain unreacted. This depends on the availability of reactants relative to their stoichiometric requirements [Per the Limiting Reagent Principle].

Step 4: Logical Conclusion based on Reactant Availability

If the reactants are provided in any ratio other than exactly $1:3$, one reactant will inevitably remain unreacted. In a real-world system, unless the reactants are introduced in precise stoichiometric proportions ($1 \text{ mole } N_2$ to $3 \text{ moles } H_2$), one reactant will be consumed completely (Limiting Reagent) while the other will remain in excess [Law of Definite Proportions].

Final Solution: Yes, one of the reactants will remain unreacted if the initial amounts provided are not in the exact stoichiometric ratio of 1:3. If the ratio of moles of $H_2$ to $N_2$ is less than 3, $N_2$ remains; if the ratio is greater than 3, $H_2$ remains.

Step 1: Understanding Scientific Notation Requirements

Scientific notation represents a number in the form $N \times 10^n$, where $N$ is a coefficient such that $1 \le |N| < 10$ and $n$ is an integer exponent [Per the International System of Units (SI) standards for numerical representation].

Step 2: Identifying the Significant Figures

In the integer $8008$, all digits are significant. The zeros are trapped between non-zero digits, which characterizes them as significant according to the rules of significant figures [Rule: Zeros between non-zero digits are always significant].

Step 3: Calculating the Decimal Shift

To express $8008$ in the form $N \times 10^n$, we must place the decimal point after the first non-zero digit.
Original number: $8008.0$
Target placement: $8.008$
To move the decimal point three places to the left, we divide by $10^3$: $8008 \div 1000 = 8.008$ [By the Law of Indices, to maintain equality, we must multiply the result by $10^3$].

Step 4: Final Formalization

Combining the coefficient and the power of ten: $8.008 \times 10^3$

Final Solution: 8.008 \times 10^3

Step 1: Determining Molar Masses

To determine the mass of copper (Cu) in $CuSO_4$, we must first calculate the molar mass of the compound based on the atomic masses of the constituent elements [Per the Law of Definite Proportions].

• Atomic mass of Copper ($Cu$) = $63.5\text{ g/mol}$
• Atomic mass of Sulphur ($S$) = $32.0\text{ g/mol}$
• Atomic mass of Oxygen ($O$) = $16.0\text{ g/mol}$

Molar mass of $CuSO_4$ = $(1 \times 63.5) + (1 \times 32.0) + (4 \times 16.0)$

Molar mass of $CuSO_4$ = $63.5 + 32.0 + 64.0 = 159.5\text{ g/mol}$

Step 2: Establishing Mass Fraction

The mass fraction of an element in a compound is calculated by dividing the total mass of that element in the formula by the total molar mass of the compound.

Mass fraction of $Cu$ = $\frac{\text{Mass of } Cu}{\text{Molar mass of } CuSO_4}$

Mass fraction of $Cu$ = $\frac{63.5}{159.5} \approx 0.3981$

Step 3: Calculating Mass in 100g of Sample

To find the mass of copper in 100g of copper sulphate, multiply the total mass of the sample by the mass fraction of copper [By the principle of conservation of mass].

Mass of $Cu$ = $\text{Total mass of } CuSO_4 \times \text{Mass fraction of } Cu$

Mass of $Cu = 100\text{ g} \times \left( \frac{63.5}{159.5} \right)$

Mass of $Cu = \frac{6350}{159.5} \approx 39.81\text{ g}$

Summary of Data

Final Solution: 39.81 g of copper can be obtained from 100 g of copper sulphate.

Step 1: Analyzing the Balanced Chemical Equation

The given chemical reaction is:

$A + B_2 \rightarrow AB_2$

According to the stoichiometric coefficients in the balanced equation [Law of Conservation of Mass]:

• $1 \text{ mole of } A$ reacts with $1 \text{ mole of } B_2$.
• The product formed is $1 \text{ mole of } AB_2$.

Step 2: Assessing the Reaction Mixture

Given the reaction mixture:

• Initial moles of $A = 2 \text{ mol}$
• Initial moles of $B = 3 \text{ mol}$

Note: The reactant $B$ is provided as $B$ (atomic form) in the question prompt, but the balanced equation consumes $B_2$ (molecular form). Assuming the intended reactant is $B_2$ based on the stoichiometry of $AB_2$, we evaluate the consumption based on the balanced equation $A + B_2 \rightarrow AB_2$.

Step 3: Calculating Required Amounts and Determining Limiting Reagent

To determine the limiting reagent, we compare the stoichiometric requirement with the actual availability [Stoichiometric Principle]:

By comparing the ratios:

For $A$: $\frac{\text{Available}}{\text{Stoichiometric Coefficient}} = \frac{2}{1} = 2$

For $B_2$: $\frac{\text{Available}}{\text{Stoichiometric Coefficient}} = \frac{3}{1} = 3$

Since the ratio for $A$ ($2$) is less than the ratio for $B_2$ ($3$), $A$ will be consumed entirely before $B_2$ is exhausted [Definition of Limiting Reagent].

Step 4: Conclusion

Because $A$ is completely consumed, it limits the extent of the reaction and determines the maximum amount of product $AB_2$ that can be formed. $B_2$ is present in excess ($3 \text{ mol} - 2 \text{ mol} = 1 \text{ mol}$ remains unreacted).

Final Solution: The limiting reagent is A.

Step 1: Analyzing the Numerical Value

The given number is $500.0$. To determine the number of significant figures, we must apply the standard rules of measurement and precision established in the International System of Units (SI) and analytical chemistry guidelines.

Step 2: Theoretical Justification (Rules for Significant Figures)

• Rule for Trailing Zeros: Zeros to the right of a decimal point are considered significant if they are part of a measured value, as they indicate the precision of the instrument used [Per IUPAC Guidelines for Significant Figures].
• Rule for Non-Zero Digits: All non-zero digits are inherently significant [Axiom of Measured Data].
• Rule for Zeros between Significant Digits: Zeros occurring between two non-zero digits are always significant, regardless of the decimal placement [Standard Convention of Precision].

Step 3: Sequential Breakdown of 500.0

Step 4: Logical Synthesis

In the number $500.0$, the presence of the decimal point indicates that the measurement is precise to the tenths place. Therefore, the trailing zero is not merely a placeholder but a reflection of the instrument's limit of detection. All four digits ($5, 0, 0, 0$) contribute to the reported precision of the measurement.

Final Solution: The number 500.0 has 4 significant figures.

Step 1: Establishing the Chemical Equation

The combustion of carbon in the presence of dioxygen ($O_2$) is represented by the following balanced stoichiometric chemical equation:

$C(s) + O_2(g) \rightarrow CO_2(g)$

[Per the Law of Conservation of Mass, the equation must contain an equal number of atoms for each element on both sides of the reaction.]

Step 2: Stoichiometric Analysis

Based on the balanced equation, the stoichiometric coefficients indicate the molar ratio of the reactants and products:

• $1$ mole of Carbon ($C$) reacts with $1$ mole of Dioxygen ($O_2$) to produce $1$ mole of Carbon Dioxide ($CO_2$).

[By the Law of Definite Proportions, the molar relationship between reactants and products is fixed.]

Step 3: Calculating Molar Mass of Product ($CO_2$)

To determine the mass of $CO_2$ produced, we calculate the molar mass of $CO_2$ using the standard atomic masses of Carbon ($12.01 \, g/mol$) and Oxygen ($16.00 \, g/mol$):

$Molar \, Mass \, of \, CO_2 = (1 \times Atomic \, Mass \, of \, C) + (2 \times Atomic \, Mass \, of \, O)$

$Molar \, Mass \, of \, CO_2 = (1 \times 12.01 \, g/mol) + (2 \times 16.00 \, g/mol)$

$Molar \, Mass \, of \, CO_2 = 12.01 \, g/mol + 32.00 \, g/mol = 44.01 \, g/mol$

Step 4: Visualization of the Stoichiometric Conversion

Step 5: Final Calculation

Since the reaction stoichiometry dictates that $1$ mole of $C$ produces $1$ mole of $CO_2$, and the molar mass of $CO_2$ is $44.01 \, g/mol$:

$Mass \, of \, CO_2 = Number \, of \, moles \times Molar \, Mass$

$Mass \, of \, CO_2 = 1 \, mol \times 44.01 \, g/mol = 44.01 \, g$

Final Solution: The amount of carbon dioxide produced is 44.01 g.