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 determine how many grams of $ \text{CO}_2 $ are produced from the complete combustion of 10 g of $ \text{C}_2\text{H}_6 $ (ethane), we follow these steps:

Step 1: Determine the molar mass of $ \text{C}_2\text{H}_6 $ and $ \text{CO}_2 $.

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

Step 2: Calculate the moles of $ \text{C}_2\text{H}_6 $ 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: Use the stoichiometry of the reaction to find moles of $ \text{CO}_2 $ produced.

The balanced reaction 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 of $ \text{C}_2\text{H}_6 $ to $ \text{CO}_2 $ is 2:4 or 1:2. Therefore, 0.332 mol of $ \text{C}_2\text{H}_6 $ will produce 0.664 mol of $ \text{CO}_2 $.

Step 4: Calculate the grams of $ \text{CO}_2 $ produced.

Mass of $ \text{CO}_2 $ = 0.664 mol x 44.01 g/mol = 29.23 g

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