Question

11) If 352 moles of butane (CH) is mixed with 8.47 moles of oxygen gas, how many grams of water will be produced?

2C,H,0 + 130, 800, + 10 H,0 (balanced)

Answer 1

117 g

Step-by-step explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 63.55 32.00 18.02

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol: 352 8.47

2. Calculate the amount of water that each reactant can produce

(a) From C₄H₁₀

Moles of H₂O = 352 mol C₄H₁₀ × (10 mol H₂O/2 mol C₄H₁₀) = 704 mol H₂O

(b) From O₂:

Moles of H₂O = 8.47 mol O₂ × (10 mol H₂O/13 mol O₂) = 6.515 mol H₂O

3. Identify the limiting reactant

The limiting reactant is O₂, because it gives fewer moles of H₂O.

4. Calculate the mass of water

Mass of H₂ = 5.515 mol H₂O × (18.02 g H₂)/(1 mol H₂O) = 117 g H₂O

The reaction produces 117 g H₂O.