The reaction for combustion of hydrogen gas in the oxygen is as follows:
[tex]2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(g)[/tex]
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 [tex]H_{2}[/tex] is 2 g/mol converting mass into number of moles:
[tex]n=\frac{m}{M}=\frac{1.80 g}{2 g/mol}=0.9 mol[/tex]
Thus, 0.9 mol of [tex]H_{2}[/tex] gives 0.9 mol of [tex]H_{2}O(g)[/tex]. Molar mass of [tex]H_{2}O(g)[/tex] 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 [tex]H_{2}O(g)[/tex] produced will be 16.2 g.
