Question

How many grams of H2 are required to completely convert 80g of Fe2O3?

Answer 1

3.0 grams H₂ is required.

Step-by-step explanation

Given:

Mass of Fe2O3 = 80 g

Equation:

What to find:

The grams of H2 required to completely convert 80g of Fe2O3.

Step-by-step solution:

From the equation of reaction;

3 moles of H2 completely react with 1 mole of Fe2O3

Note: Molar mass of H2 is 2.016 grams per mole and Molar mass of Fe2O3 is 159.69 g/mol

This implies; (3 x 2.016 g) = 6.048 grams H2 completely react with 159.69 grams Fe2O3.

Therefore, x grams H2 will completely convert 80 grams Fe2O3.

Cross multiply and divide both sides by 159.69 grams Fe2O3.

x grams H2 is now equal to

`x=\frac{80\text{ }g\text{ }Fe_2O_3}{159.69\text{ }g\text{ }Fe_2O_3}*6.048\text{ }g\text{ }H_2=3.0298\approx3.0\text{ }grams\text{ }H_2`

Therefore the grams of H2 required to completely convert 80g of Fe2O3 is 3.0 grams