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Anton [14]
3 years ago
11

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

Chemistry
2 answers:
Hoochie [10]3 years ago
8 0

<u>Answer:</u> The theoretical yield of water is 162 grams.

<u>Explanation:</u>

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.

Lapatulllka [165]3 years ago
5 0
The theoretical yield of H2O if 130g of H2O is produced from 18g of H2 and an excess of O2 is 160g. 
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