Questions: The gaseous hydrocarbon acetylene, C2H2, used in welders' torches, burns according to the following equation: 2 C2H2(g) + 5 O2(g) -> 4 CO2(g) + 2 H2O(g) What is the theoretical yield, in grams, of CO2, if 22.0 g of C2H2 completely reacts? Express your answer with the appropriate units.

Solution Steps

The molar mass of acetylene (\(\mathrm{C}_2\mathrm{H}_2\)) is calculated as follows:

• Carbon (\(\mathrm{C}\)): \(12.01 \, \text{g/mol}\)
• Hydrogen (\(\mathrm{H}\)): \(1.008 \, \text{g/mol}\)

\[ \text{Molar mass of } \mathrm{C}_2\mathrm{H}_2 = 2 \times 12.01 + 2 \times 1.008 = 26.036 \, \text{g/mol} \]

Given 22.0 g of \(\mathrm{C}_2\mathrm{H}_2\), we calculate the moles:

\[ \text{Moles of } \mathrm{C}_2\mathrm{H}_2 = \frac{22.0 \, \text{g}}{26.036 \, \text{g/mol}} = 0.8449 \, \text{mol} \]

From the balanced chemical equation:

\[ 2 \mathrm{C}_2\mathrm{H}_2 + 5 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 2 \mathrm{H}_2\mathrm{O} \]

The mole ratio of \(\mathrm{C}_2\mathrm{H}_2\) to \(\mathrm{CO}_2\) is 2:4, or 1:2. Therefore, the moles of \(\mathrm{CO}_2\) produced are:

\[ \text{Moles of } \mathrm{CO}_2 = 0.8449 \, \text{mol} \times 2 = 1.6898 \, \text{mol} \]

The molar mass of \(\mathrm{CO}_2\) is:

• Carbon (\(\mathrm{C}\)): \(12.01 \, \text{g/mol}\)
• Oxygen (\(\mathrm{O}\)): \(16.00 \, \text{g/mol}\)

\[ \text{Molar mass of } \mathrm{CO}_2 = 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} \]

The theoretical yield in grams is:

\[ \text{Theoretical yield of } \mathrm{CO}_2 = 1.6898 \, \text{mol} \times 44.01 \, \text{g/mol} = 74.347 \, \text{g} \]

Final Answer