Questions: Question 8 1 pts Assuming complete combustion, how many grams of butane (C4H10) would be required to consume 32.0 g O2 ? (Give the balanced chemical equation) 1.82 g 4.47 g 16.6 g 8.94 g 8.30 g

Question 8
1 pts

Assuming complete combustion, how many grams of butane (C4H10) would be required to consume 32.0 g O2 ?
(Give the balanced chemical equation)
1.82 g
4.47 g
16.6 g
8.94 g
8.30 g

Transcript text: Question 8 1 pts Assuming complete combustion, how many grams of butane $\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)$ would be required to consume 32.0 g $\mathrm{O}_{2}$ ? (Give the balanced chemical equation) 1.82 g 4.47 g 16.6 g 8.94 g 8.30 g Previous Next

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the complete combustion of butane ($\mathrm{C}_4\mathrm{H}_{10}$) is:

\[ 2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O} \]

Step 2: Calculate Moles of $\mathrm{O}_2$

First, calculate the number of moles of $\mathrm{O}_2$ in 32.0 g:

\[ \text{Molar mass of } \mathrm{O}_2 = 32.00 \, \text{g/mol} \]

\[ \text{Moles of } \mathrm{O}_2 = \frac{32.0 \, \text{g}}{32.00 \, \text{g/mol}} = 1.000 \, \text{mol} \]

Step 3: Use Stoichiometry to Find Moles of $\mathrm{C}_4\mathrm{H}_{10}$

From the balanced equation, 13 moles of $\mathrm{O}_2$ react with 2 moles of $\mathrm{C}_4\mathrm{H}_{10}$. Therefore, the moles of $\mathrm{C}_4\mathrm{H}_{10}$ required are:

\[ \text{Moles of } \mathrm{C}_4\mathrm{H}_{10} = \frac{2}{13} \times 1.000 \, \text{mol} = 0.1538 \, \text{mol} \]

Step 4: Convert Moles of $\mathrm{C}_4\mathrm{H}_{10}$ to Grams

Calculate the mass of $\mathrm{C}_4\mathrm{H}_{10}$ needed:

\[ \text{Molar mass of } \mathrm{C}_4\mathrm{H}_{10} = 58.12 \, \text{g/mol} \]

\[ \text{Mass of } \mathrm{C}_4\mathrm{H}_{10} = 0.1538 \, \text{mol} \times 58.12 \, \text{g/mol} = 8.94 \, \text{g} \]