Question

____ $c_3h_8$ + _____$o_2$ $rightarrow$ _____ $co_2$ + ____ $h_2o$
if 85g of water is produced how many grams of propane are combusted?(20 points)

Explanation:

Step1: Balance the chemical equation

The unbalanced equation is \( \text{C}_3\text{H}_8 + \text{O}_2
ightarrow \text{CO}_2 + \text{H}_2\text{O} \).

• For carbon: There are 3 C in \( \text{C}_3\text{H}_8 \), so we put 3 in front of \( \text{CO}_2 \), giving \( \text{C}_3\text{H}_8 + \text{O}_2

ightarrow 3\text{CO}_2 + \text{H}_2\text{O} \).

• For hydrogen: There are 8 H in \( \text{C}_3\text{H}_8 \), so we put 4 in front of \( \text{H}_2\text{O} \) (since \( 4\times2 = 8 \)), giving \( \text{C}_3\text{H}_8 + \text{O}_2

ightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \).

• For oxygen: On the right, we have \( 3\times2 + 4\times1 = 10 \) O. So we put 5 in front of \( \text{O}_2 \) (since \( 5\times2 = 10 \)). The balanced equation is \( \text{C}_3\text{H}_8 + 5\text{O}_2

ightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \).

Step2: Calculate moles of \( \text{H}_2\text{O} \)

The molar mass of \( \text{H}_2\text{O} \) is \( (2\times1) + 16 = 18 \, \text{g/mol} \).
Given mass of \( \text{H}_2\text{O} = 85 \, \text{g} \).
Moles of \( \text{H}_2\text{O} = \frac{\text{mass}}{\text{molar mass}} = \frac{85}{18} \approx 4.722 \, \text{mol} \).

Step3: Use mole ratio from balanced equation

From the balanced equation, the mole ratio of \( \text{C}_3\text{H}_8 \) to \( \text{H}_2\text{O} \) is \( 1:4 \).
Let moles of \( \text{C}_3\text{H}_8 \) be \( x \). Then \( \frac{x}{4.722} = \frac{1}{4} \), so \( x = \frac{4.722}{4} \approx 1.1805 \, \text{mol} \).

Step4: Calculate mass of \( \text{C}_3\text{H}_8 \)

Molar mass of \( \text{C}_3\text{H}_8 \) is \( (3\times12) + (8\times1) = 36 + 8 = 44 \, \text{g/mol} \).
Mass of \( \text{C}_3\text{H}_8 = \text{moles} \times \text{molar mass} = 1.1805 \times 44 \approx 51.94 \, \text{g} \).

Answer:

Approximately \( 51.9 \, \text{g} \) (or more precisely \( \approx 52 \, \text{g} \)) of propane are combusted.