Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is ___________ × 10⁻¹ g. (in nearest integer)

Solution

Chemistry ma’am · 5 steps

Balance the equation first; then each mole of butane gives 5 moles of water, and the rest is counting.

1. Balanced equation. Butane burns completely to carbon dioxide and water. Balancing C, then H, then O gives the equation below. Per mole of butane, that is 5 moles of water.

   2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

2. Moles of water. Molar mass of H₂O is 18 g mol⁻¹, so 72.0 g is 72/18 = 4 mol.
3. Moles of butane. From the ratio 2 : 10, butane needed = 4 × 2/10 = 0.8 mol.
4. Mass of butane. Molar mass of C₄H₁₀ is 4 × 12 + 10 × 1 = 58 g mol⁻¹, so the mass is 0.8 × 58 = 46.4 g.
5. So. 46.4 g = 464 × 10⁻¹ g, so the answer is 464.

The common slip: 46. Reporting the mass in grams gives 46, but the blank is in units of 10⁻¹ g, so 46.4 g must be written as 464.

The idea to keep. In any stoichiometry question, go grams to moles, use the mole ratio from the balanced equation, then go back to grams. Always check the unit the blank asks for.

Answer from the NTA final answer key. Solution by PrepWiser.