Question

How many grams of $ \text{CO}_2 $ are produced when 10 g of $ \text{C}_2\text{H}_6 $ (ethane) is completely combusted? Reaction: $ 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} $

• A. 29.3 g
• B. 44.0 g
• C. 58.6 g
• D. 88.0 g

Hint:

To solve stoichiometry problems involving mass-to-mass conversions: \begin{enumerate} \item \textbf{Balance the Chemical Equation:} Ensure the reaction is correctly balanced to determine the correct mole ratios between reactants and products. \item \textbf{Calculate Molar Masses:} Determine the molar mass of the given substance and the substance you need to find. \item \textbf{Convert Given Mass to Moles:} Use the molar mass to convert the given mass of the substance into moles. \item \textbf{Use Mole Ratio (Stoichiometry):} From the balanced equation, use the stoichiometric coefficients to find the mole ratio between the given substance and the desired substance. Use this ratio to convert moles of the given substance to moles of the desired substance. \item \textbf{Convert Moles to Desired Mass:} Use the molar mass of the desired substance to convert its moles back into grams. \end{enumerate}

Answer 1

The Correct Option is A

To ascertain the mass of \( \text{CO}_2 \) generated from the complete combustion of 10 g of \( \text{C}_2\text{H}_6 \) (ethane), the following procedure is employed:

Step 1: Determine the molar masses of \( \text{C}_2\text{H}_6 \) and \( \text{CO}_2 \).

• Molar mass of \( \text{C}_2\text{H}_6 \): (2 × 12.01 g/mol for C) + (6 × 1.008 g/mol for H) = 30.07 g/mol
• Molar mass of \( \text{CO}_2 \): (1 × 12.01 g/mol for C) + (2 × 16.00 g/mol for O) = 44.01 g/mol

Step 2: Compute the molar quantity of \( \text{C}_2\text{H}_6 \) present in 10 g.

Moles of \( \text{C}_2\text{H}_6 \) = \( \frac{10 \text{ g}}{30.07 \text{ g/mol}} \) ≈ 0.332 mol

Step 3: Utilize the reaction stoichiometry to determine the moles of \( \text{CO}_2 \) produced.

The balanced chemical equation is: \( 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \)

The molar ratio between \( \text{C}_2\text{H}_6 \) and \( \text{CO}_2 \) is 2:4, simplifying to 1:2. Consequently, 0.332 mol of \( \text{C}_2\text{H}_6 \) yields 0.664 mol of \( \text{CO}_2 \).

Step 4: Calculate the mass of \( \text{CO}_2 \) generated.

Mass of \( \text{CO}_2 \) = 0.664 mol × 44.01 g/mol = 29.23 g

Therefore, the mass of \( \text{CO}_2 \) produced is approximately '29.3 g'.