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=(m)/(M)=(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.