How many grams of solid aluminum sulfide can be prepared by the reaction of 10.0 g of aluminum and 15.0 g of sulfur? How many grams of the nonlimiting reactant are in excess?

Butane ($\text{C}_4\text{H}_{10}$) burns in oxygen ($\text{O}_2$) to produce carbon dioxide ($\text{CO}_2$) and water vapor ($\text{H}_2\text{O}$).

• Molar mass of $\text{C}_4\text{H}_{10} = (4 \times 12.01) + (10 \times 1.008) = 48.04 + 10.08 = 58.12 \text{ g/mol}$
• Molar mass of $\text{O}_2 = 2 \times 16.00 = 32.00 \text{ g/mol}$
• Molar mass of $\text{CO}_2 = 12.01 + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \text{ g/mol}$

Given: $4.65 \text{ g}$ of butane and $40.0 \text{ g}$ of oxygen.

Moles of $\text{C}_4\text{H}_{10}$:

Moles of $\text{O}_2$:

Using the stoichiometric coefficients from the balanced equation:
For $\text{C}_4\text{H}_{10}$: $0.07999 \text{ mol C}_4\text{H}_{10} \times \frac{13 \text{ mol O}_2}{2 \text{ mol C}_4\text{H}_{10}} = 0.5199 \text{ mol O}_2$ needed.
Since we have $1.25 \text{ mol O}_2$, which is more than $0.5199 \text{ mol O}_2$ needed, $\text{C}_4\text{H}_{10}$ is the limiting reactant.
$\text{O}_2$ is the excess reactant.

From the balanced equation, $2 \text{ mol C}_4\text{H}_{10}$ produces $8 \text{ mol CO}_2$.

Moles of $\text{CO}_2$ produced:

Mass of $\text{CO}_2$ produced:

The mass of carbon dioxide produced is $\boxed{\text{14.08 g}}$.

Moles of $\text{O}_2$ consumed:

Moles of $\text{O}_2$ remaining:

Mass of $\text{O}_2$ remaining:

The mass of excess reactant remaining is $\boxed{\text{23.36 g}}$.