Question

If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?

Answer 1

Answer:  0.745 g of $H_2SO_4$ will be produced from  1.08 g of sodium sulfate

Explanation:

To calculate the moles :

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

$\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles$  

$3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4$

$Na_2SO_4$ is the limiting reagent as it limits the formation of product and $H_3PO_4$ is the excess reagent.

According to stoichiometry :

3 moles of $Na_2SO_4$ produce = 3 moles of $H_2SO_4$

Thus 0.0076 moles of $Na_2SO_4$ will require=$\frac{3}{3}\times 0.0076=0.0076moles$  of $H_2SO_4$

Mass of $H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g$

Thus 0.745 g of $H_2SO_4$ will be produced from  1.08 g of sodium sulfate