Question

A hydrogen-filled balloon was ignited and 1.80 g of hydrogen reacted with 14.4 g of oxygen. part a how many grams of water vapor were formed? (assume that water vapor is the only product.)

Answer 1

The reaction for combustion of hydrogen gas in the oxygen is as follows:

$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)$

Here, oxygen is present in excess thus, hydrogen will be limiting reactant.

2 moles of hydrogen gives 2 moles of water vapor, thus, 1 mole will give 1 mole of water vapor.

The mass of hydrogen is 1.80 g and molar mass of $H_{2}$ is 2 g/mol converting mass into number of moles:

$n=\frac{m}{M}=\frac{1.80 g}{2 g/mol}=0.9 mol$

Thus, 0.9 mol of $H_{2}$ gives 0.9 mol of $H_{2}O(g)$. Molar mass of $H_{2}O(g)$ is 18 g/mol, converting number of moles to mass,

m=n×M=0.9 mol× 18 g/mol=16.2 g

Therefore, mass of water vapor $H_{2}O(g)$ produced will be 16.2 g.