Question

When 0.125 mol butane, C₄H₁₀, is burned with excess oxygen, ideally how many moles of carbon dioxide are formed?

(2)C₄H₁₀ (g) + (13) O₂ (g) => 8 CO₂ (g) + (10)H₂0 (l)

a) 0.250 mol
b) 0.375 mol
c) 0.500 mol
d) 0.750 mol

Answer 1

Burning 0.125 mol of butane with excess oxygen produces c)0.500 mol of carbon dioxide, determined by using the stoichiometry of the balanced chemical equation for the combustion of butane.

Step-by-step explanation:

When 0.125 mol butane (C4H10) is burned with excess oxygen, the balanced chemical equation is 2C4H10 (g) + 13O2 (g) → 8CO2 (g) + 10H2O (l).

From this equation, we can see that 2 moles of butane produce 8 moles of carbon dioxide.

Therefore, if 0.125 mol of butane is combusted, we use a simple ratio to determine the moles of CO2 formed:

2 moles C4H10 : 8 moles CO2
0.125 moles C4H10 : X moles CO2

Calculating this:

X = (0.125 moles C4H10 × 8 moles CO2) ÷ 2 moles C4H10
X = 0.5 moles CO2

So the correct answer is c) 0.500 mol of CO2 are formed when 0.125 mol of butane is burned with excess oxygen.