Butane reacts with oxygen to produce carbon dioxide and water following the equation given below.
C₄ H₁₀( g)+13/2 O₂( g) → 4 CO₂( g)+5 H₂ O(l)
If 174.0 kg of butane is mixed with 320.0 kg of O₂, the volume of water formed in liters is
_____________. (Nearest integer)
[Given : (a) Molar mass of C, H, O are 12,1,16 g mol⁻¹ respectively, (b) Density of water =1 g mL⁻¹]

Solution

Chemistry ma’am · 4 steps

Do not trust the bigger mass: check how much oxygen the butane needs, and oxygen runs out first.

1. Convert to moles. Molar mass of C₄H₁₀ is 48 + 10 = 58 g mol⁻¹, so 174000 g is 3000 mol. For O₂ (32 g mol⁻¹), 320000 g is 10000 mol.
2. Find the limiting reagent. Each mole of butane needs 13/2 = 6.5 mol of O₂. So 3000 mol of butane would need 3000 × 6.5 = 19500 mol of O₂, but only 10000 mol is present. Oxygen is the limiting reagent, and butane is left over.

   n_(H₂O) = 10000 × 5/6.5 = 7692.3 mol

3. Mass to volume. Mass of water is 7692.3 × 18 = 138461.5 g. With a density of 1 g mL⁻¹, that is 138461.5 mL = 138.46 L.
4. So. The volume of water formed is about 138.46 L, which rounds to 138.

The common slip: 270. Taking butane as the limiting reagent gives 3000 × 5 = 15000 mol of water, or 270 L, not 138. Oxygen runs out long before the butane does.

The idea to keep. To find the limiting reagent, divide the moles of each reactant by its coefficient in the balanced equation. The smaller result runs out first, and every product is calculated from that reactant.

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