Question

NH₄NO₃ → N₂O + 2H₂O When 45.70 g of NH₄NO₃ decomposes, what mass of each product is formed?

Answer 1

Answer: 25.13 g of $N_2O$  and 20.56 g of $H_2O$ will be produced from 45.70 g of $NH_4NO_3$

Explanation:

To calculate the moles :

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

$\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles$

The balanced chemical equation is:

$NH_4NO_3\rightarrow N_2O+2H_2O$  

According to stoichiometry :

1 mole of $NH_4NO_3$ produce = 1 mole of $N_2O$

Thus 0.571 moles of $NH_4NO_3$ will require=$\frac{1}{1}\times 0.571=0.571moles$  of $N_2O$  

Mass of $N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g$

1 mole of $NH_4NO_3$ produce = 2 moles of $H_2O$

Thus 0.571 moles of $NH_4NO_3$ will require=$\frac{2}{1}\times 0.571=1.142moles$  of $H_2O$  

Mass of $H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g$

Thus 25.13 g of $N_2O$  and 20.56 g of $H_2O$ will be produced from 45.70 g of $NH_4NO_3$