Question

acetylene gas (c2h2) burna in the presence of oxygen gas to produce carbon dioxide gas and water vapor. a reaction produces 20.1 g of carbon dioxide gas. how many moles of acetylene must have been combusted

Answer 1

To determine how many moles of acetylene (C₂H₂) were combusted to produce 20.1 grams of carbon dioxide (CO₂), we first need to understand the balanced chemical equation for the combustion of acetylene:

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

From the balanced equation, we can see that 2 moles of C₂H₂ produce 4 moles of CO₂. This means that:

- 1 mole of C₂H₂ produces 2 moles of CO₂.

Next, we will calculate the number of moles of CO₂ produced from the given mass (20.1 grams).

1. Calculate the molar mass of CO₂:
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol, so for CO₂:
\[
\text{Molar mass of CO}_2 = 12.01 + 2(16.00) = 12.01 + 32.00 = 44.01 \, \text{g/mol}
\]

2. Calculate the number of moles of CO₂ produced:
\[
\text{Moles of CO}_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} = \frac{20.1 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.456 \, \text{moles of CO}_2
\]

3. Use the stoichiometry from the balanced equation to find the moles of C₂H₂:
According to the balanced equation:
\[
2 \, \text{C}_2\text{H}_2 \text{ produces } 4 \, \text{CO}_2
\]
Therefore:
\[
1 \, \text{C}_2\text{H}_2 \text{ produces } 2 \, \text{CO}_2
\]
This implies that:
\[
\text{Moles of C}_2\text{H}_2 = \frac{1}{2} \times \text{Moles of CO}_2 = \frac{1}{2} \times 0.456 \approx 0.228 \, \text{moles of C}_2\text{H}_2
\]

Thus, approximately **0.228 moles of acetylene** were combusted to produce 20.1 grams of carbon dioxide gas.