Question

When 27.50 liters of oxygen gas reacts with excess hydrogen gas at STP how many grams of water is produced?

Answer 1

Answer: The mass of water produced is 44.28 g

Step-by-step explanation:

We are given:

Volume of oxygen gas = 27.50 L

At STP conditions:

22.4 L of volume is occupied by 1 mole of a substance

27.50 L of oxygen gas will be occupied by =
`(1mol)/(22.4L)* 27.50L=1.23mol`

The chemical equation for the formation of water follows:

`2H_2+O_2\rightarrow 2H_2O`

By the stoichiometry of the reaction:

If 1 mole of oxygen gas produces 2 moles of water

So, 1.23 moles of oxygen gas will produce =
`(2)/(1)* 1.23=2.46mol` of water

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

Molar mass of water = 18 g/mol

Plugging values in equation 1:

`\text{Mass of water}=(2.46mol* 18g/mol)=44.28g`

Hence, the mass of water produced is 44.28 g