Consider the following reaction

Question

Consider the following reaction:

2c4h10(g)+13o2(g)→8co2(g)+10h2o(g)

Suppose you are given 3.0 moles of Butane, C10. How many moles of O2 would you need?

Answer 1

2 C₄H₁₀ (g) + 13 O₂(g) → 8 CO₂ (g) + 10 H₂O (g)

To solve this problem, we need to first determine the mole ratio between butane and oxygen from the balanced chemical reaction. From above, you can see that 13 moles of O₂ reacts with 2 moles of C₄H₁₀. From this, you can write 2 conversion factors:

(13 mol O₂ / 2 mol C₄H₁₀) OR (2 mol C₄H₁₀ / 13 mol O₂)

We are starting with 3.0 moles of C₄H₁₀. Which conversion factor allows you to cancel moles of C₄H₁₀ and end up with moles of O₂? The first one!! The stoichiometry conversion can then be written as:

3.0 mol C₄H₁₀ x (13 mol O₂ / 2 mol C₄H₁₀) = 19.5 mol O₂ ≅ 20. mol O₂ (rounded to 2 significant figures)

Answer 2

Based on the balanced chemical equation you have, for every 2 moles of butane, 13 moles of oxygen gas react with it. Thus, the molar ratio between the two is 2:13. Given that we have 3 moles of butane, we can determine the moles of oxygen based on the molar ratio:

13:2 = x:3

3*(13:2) = x

x = 39:2

x = 19.5 mol O₂