Question

C3H8 + 5 O2 arrow 3 CO2 + 4 H2O

Calculate the mass of H2O is produced from the reaction of 6.3 g of propane.

Answer 1

In the reaction

C₃H₈ + 5 O₂ $\rightarrow$ 3 CO₂ + 4 H₂O

1 mol of propane produces 4 mol of water:

$n(C_3H_8) = \frac{n(H_2O)}{4}$

$n(H_2O) = 4·n(C_3H_8)$ .

The number of the moles of propane is the ratio of its mass $m$ to its molar mass $M$ (44.10 g/mol):

$n(C_3H_8) = \frac{m}{M} = \frac{6.3(g)}{44.10 (g/mol)} = 0.143 (mol)$ .

Making use of the same relation to calculate the mass of water produced ($M = 18.02$ g/mol):

$m = nM = 4·0.143·18.02 = 10.3$ g.

Answer: 10.3 g of H₂O is produced from the reaction of 6.3 g of propane.