Question

2Fe(OH)3 Fe2O3+3H2O how many grams of Fe2O3 are produced if 10.7 grams of Fe (OH)3 react in this way

Answer 1

Answer: 7.98 grams of $Fe_2O_3$ are produced if 10.7 grams of $Fe(OH)_3$ are reacted.

Explanation:

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

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

Putting values in equation 1, we get:

$\text{Moles of} Fe(OH)_3=\frac{10.7g}{106.87g/mol}=0.100mol$  

The chemical equation for the reaction is

$2Fe(OH)_3\rightarrow Fe_2O_3+3H_2O$

By Stoichiometry of the reaction:

2 moles of $Fe(OH)_3$ produce = 1 mole of $Fe_2O_3$

So, 0.100 moles of $Fe(OH)_3$ produce= $\frac{1}{2}\times 0.100=0.05mol$ of $Fe_2O_3$  

Mass of $Fe_2O_3$ =$moles\times {\text{Molar Mass}}=0.05mol\times 159.69g/mol=7.98g$  

Hence 7.98 grams of $Fe_2O_3$ are produced if 10.7 grams of $Fe(OH)_3$ are reacted.