Question

What is the theoretical yield of H2O if 130 g of H2O is produced from 18 g of H2 and an excess of O2?

Answer 1

Answer: The theoretical yield of water is 162 grams.

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of hydrogen gas = 18 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{18g}{2g/mol}=9mol$

The chemical equation for the formation of water from hydrogen gas and oxygen gas follows:

$2H_2+O_2\rightarrow 2H_2O$

As, oxygen gas is present in excess. It is considered as an excess reagent.

So, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 2 moles of water

So, 9 moles of hydrogen gas will produce = $\frac{2}{2}\times 9=9mol$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 9 moles

Putting values in equation 1, we get:

$9mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=162g$

Hence, the theoretical yield of water is 162 grams.

Answer 2

The theoretical yield of H2O if 130g of H2O is produced from 18g of H2 and an excess of O2 is 160g.