Question

Balanced chemical equation: 2C8H18(g)+25O2(g)->16CO2(g)+18H2O(g)

0.320 mol of octane is allowed to react with 0.880 mol of oxygen. How many moles of water are produced in this reaction?

Answer 1

We first need to determine which is the limiting reactant. For this, we divide the number fo moles of each species by its corresponding stoichiometric coefficient and, the one with the lowest result, is the limiting reactant:

For octane:

$\frac{0.320 moles}{2}=0.160moles$

For oxygen:

$\frac{0.880 moles}{25}=0.0352moles$

Oxygen is the limiting reactant.

We know that 25 moles of O₂ produce 18 moles of water, so the moles of water that will be produced from 0.880 moles of oxygen are:

$n_{water}=0.880 moles_{O_{2}}\cdot \frac{18moles_{H_{2}O}}{25moles_{O_{2}}}=0.634moles_{H_{2}O}$