Question

c₃h₈(g) + 5o₂(g) → 3co₂(g) + 4h₂o(g)
the combustion of c₃h₈ occurs at stp.
65.0 g c₃h₈ reacts with 105 l o₂. what
is the limiting reactant?

1. c₃h₈ 2. o₂

enter the answer choice number.

remember, 1 mole gas = 22.4 l at stp.

Explanation:

Step1: Calculate moles of \( C_3H_8 \)

Molar mass of \( C_3H_8 \) is \( 3\times12.01 + 8\times1.008 = 44.094 \, \text{g/mol} \).
Moles of \( C_3H_8 = \frac{65.0 \, \text{g}}{44.094 \, \text{g/mol}} \approx 1.474 \, \text{mol} \).

Step2: Calculate moles of \( O_2 \) at STP

Using \( 1 \, \text{mol gas} = 22.4 \, \text{L} \) at STP,
Moles of \( O_2 = \frac{105 \, \text{L}}{22.4 \, \text{L/mol}} \approx 4.6875 \, \text{mol} \).

Step3: Compare required moles from reaction

Reaction: \( C_3H_8 + 5O_2
ightarrow 3CO_2 + 4H_2O \).
Moles of \( O_2 \) required for \( 1.474 \, \text{mol} \, C_3H_8 \): \( 1.474 \times 5 = 7.37 \, \text{mol} \).
We have only \( 4.6875 \, \text{mol} \, O_2 \), which is less than 7.37 mol.
Alternatively, moles of \( C_3H_8 \) required for \( 4.6875 \, \text{mol} \, O_2 \): \( \frac{4.6875}{5} = 0.9375 \, \text{mol} \), which is less than 1.474 mol.
Thus, \( O_2 \) is limiting as it runs out first.

Answer:

2