To determine the mass of [tex]\(\text{CO}_2(g)\)[/tex] produced when 2.89 L of propane [tex]\(\text{C}_3\text{H}_8(g)\)[/tex] reacts with 12.6 L of [tex]\(\text{O}_2(g)\)[/tex] at SATP, we can follow these steps:
1. Convert the given volumes to moles using the molar volume at SATP. At SATP (Standard Ambient Temperature and Pressure), 1 mole of any ideal gas occupies 24.8 L.
- Moles of propane [tex]\(\text{C}_3\text{H}_8(g)\)[/tex]:
[tex]\[
\frac{2.89 \text{ L}}{24.8 \text{ L/mol}} = 0.11653225806451613 \text{ mol}
\][/tex]
- Moles of oxygen [tex]\(\text{O}_2(g)\)[/tex]:
[tex]\[
\frac{12.6 \text{ L}}{24.8 \text{ L/mol}} = 0.5080645161290323 \text{ mol}
\][/tex]
2. Use the balanced chemical equation to identify the molar ratio of reactants to determine the limiting reagent. The balanced equation is:
[tex]\[
\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)
\][/tex]
- From the ratio, we need 5 moles of [tex]\(\text{O}_2(g)\)[/tex] for 1 mole of [tex]\(\text{C}_3\text{H}_8(g)\)[/tex].
- Calculate the required moles of [tex]\(\text{O}_2(g)\)[/tex] for the available moles of propane:
[tex]\[
0.11653225806451613 \text{ mol } \times 5 = 0.5826612903225806 \text{ mol of } \text{O}_2(g)
\][/tex]
3. Identify the limiting reagent by comparing the available [tex]\(\text{O}_2(g)\)[/tex] with the required [tex]\(\text{O}_2(g)\)[/tex]:
- Available [tex]\(\text{O}_2(g)\)[/tex] is 0.5080645161290323 mol, which is less than the required 0.5826612903225806 mol.
- Thus, [tex]\(\text{O}_2(g)\)[/tex] is the limiting reagent.
- Moles of [tex]\(\text{C}_3\text{H}_8(g)\)[/tex] that can fully react with available [tex]\(\text{O}_2(g)\)[/tex]:
[tex]\[
\frac{0.5080645161290323 \text{ mol}}{5} = 0.10161290322580645 \text{ mol of } \text{C}_3\text{H}_8(g)
\][/tex]
4. Calculate the moles of [tex]\(\text{CO}_2(g)\)[/tex] produced using the stoichiometry of the reaction:
- For 1 mole of [tex]\(\text{C}_3\text{H}_8(g)\)[/tex], 3 moles of [tex]\(\text{CO}_2(g)\)[/tex] are produced.
- Moles of [tex]\(\text{CO}_2(g)\)[/tex]:
[tex]\[
0.10161290322580645 \text{ mol} \times 3 = 0.30483870967741933 \text{ mol}
\][/tex]
5. Determine the mass of [tex]\(\text{CO}_2(g)\)[/tex] produced using its molar mass. The molar mass of [tex]\(\text{CO}_2(g)\)[/tex] is 44.01 g/mol:
- Mass of [tex]\(\text{CO}_2(g)\)[/tex]:
[tex]\[
0.30483870967741933 \text{ mol} \times 44.01 \text{ g/mol} = 13.415951612903225 \text{ g}
\][/tex]
So, the mass of [tex]\(\text{CO}_2(g)\)[/tex] produced is 13.42 g (to two decimal places).
