Question

Calculate the number of water (H2O) molecules produced from the decomposition of 75.50 grams of Iron (III) hydroxide (Fe(OH)3).

Answer 1

Answer: $6.38\times 10^{23}$ molecules of water are produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Fe(OH)_3=\frac{75.50g}{106.8g/mol}=0.707moles$

The balanced chemical reaction is:

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

According to stoichiometry :

2 moles of $Fe_2O_3$ produce = 3 moles of $H_2O$

Thus 0.707 moles of $Fe_2O_3$ will produce=$\frac{3}{2}\times 0.707=1.06moles$  of $H_2O$  

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles.

Thus 1.06 moles of $H_2O$ contains = $\frac{6.023\times 10^{23}}{1}\times 1.06=6.38\times 10^{23}$ molecules