What volume of butane (C4H10) can be produced at STP, from the reaction of 13.45 g of carbon with 17.65 L of hydrogen gas at STP?
Which reactant is in excess and how much of this reactant is left over?
4 C + 5 H2 C4H10
The Correct Answer and Explanation is:
To determine the volume of butane ((C_4H_{10})) produced from the reaction between carbon and hydrogen gas, we start with the balanced chemical equation:
[ 4C + 5H_2 \rightarrow C_4H_{10} ]
Step 1: Calculate moles of reactants
Moles of Carbon:
The molar mass of carbon ((C)) is approximately 12.01 g/mol. To find the moles of carbon in 13.45 g:
[
\text{Moles of } C = \frac{\text{mass}}{\text{molar mass}} = \frac{13.45 \, \text{g}}{12.01 \, \text{g/mol}} \approx 1.12 \, \text{mol}
]
Moles of Hydrogen:
The molar mass of hydrogen gas ((H_2)) is approximately 2.02 g/mol. To find the moles of hydrogen in 17.65 L at STP (where 1 mole of gas occupies 22.4 L):
[
\text{Moles of } H_2 = \frac{17.65 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.79 \, \text{mol}
]
Step 2: Determine the stoichiometry of the reaction
From the balanced equation:
- 4 moles of (C) react with 5 moles of (H_2).
Calculate the amount of hydrogen needed to react with the available carbon:
[
\text{Required } H_2 = 1.12 \, \text{mol } C \times \frac{5 \, \text{mol } H_2}{4 \, \text{mol } C} = 1.40 \, \text{mol } H_2
]
Step 3: Identify the limiting reactant
We have only 0.79 moles of (H_2) available, while we need 1.40 moles to react with all the carbon. Therefore, (H_2) is the limiting reactant.
Step 4: Calculate the volume of butane produced
From the stoichiometry of the reaction, 5 moles of (H_2) produce 1 mole of (C_4H_{10}). Thus, the moles of butane produced from 0.79 moles of (H_2) are:
[
\text{Moles of } C_4H_{10} = 0.79 \, \text{mol } H_2 \times \frac{1 \, \text{mol } C_4H_{10}}{5 \, \text{mol } H_2} = 0.158 \, \text{mol } C_4H_{10}
]
At STP, 1 mole of gas occupies 22.4 L, so:
[
\text{Volume of } C_4H_{10} = 0.158 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 3.54 \, \text{L}
]
Step 5: Calculate excess reactant
We started with 1.12 moles of (C). To find out how much carbon is left over after reacting with the available hydrogen:
[
\text{Moles of } C \text{ reacted} = 0.79 \, \text{mol } H_2 \times \frac{4 \, \text{mol } C}{5 \, \text{mol } H_2} = 0.632 \, \text{mol } C
]
Excess carbon:
[
\text{Excess } C = 1.12 \, \text{mol} – 0.632 \, \text{mol} \approx 0.488 \, \text{mol}
]
In grams, this is:
[
0.488 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 5.86 \, \text{g}
]
Conclusion
The reaction produces approximately 3.54 L of butane, with hydrogen as the limiting reactant. Approximately 5.86 g of carbon is left unreacted.