Question

Butane gas (C4H10) is burned to produce carbon dioxide and water vapour. If 12.8 mol of butane are completely burned, how many moles of water vapour are produced?

Answer 1

64 mol of water vapour .

Answer 2

64 mol of water vapour

Answer 3

[AI-generated answer — Explain Bot]

To find out how many moles of water vapor are produced when 12.8 mol of butane is burned, we need to look at the balanced chemical equation for the combustion of butane.

The balanced equation for the combustion of butane is:

2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

From the balanced equation, we can see that for every 2 moles of butane burned, 10 moles of water vapor are produced.

To find out how many moles of water vapor are produced from 12.8 mol of butane, we can use a proportion:

(10 mol H2O / 2 mol C4H10) = (x mol H2O / 12.8 mol C4H10)

Cross-multiplying the equation, we have:

10 mol H2O * 12.8 mol C4H10 = 2 mol C4H10 * x mol H2O

Simplifying the equation:

(10 * 12.8) / 2 = x

x = 64

Therefore, when 12.8 mol of butane is completely burned, 64 moles of water vapor are produced.