Question

Calculate the mass of $N_2{F}_4$ produced by the reaction of 2 g of ammonia and 8 g of fluorine.
$2N{H}_3+5F_2\to N_2{F}_4+6HF$
(Molar mass of fluorine = 19 g/mol)

• A. 4.368 g
• B. 10.523 g
• C. 5.682 g
• D. 15.391 g

Answer 1

The correct option is A 4.368 g

$2N{H}_3+5F_2\to N_2{F}_4+6HF$
$MolesofN{H}_3=\frac{\text{given mass}}{\text{molar mass}}=\frac{2}{17}=0.118mol\text{Moles of}{F}_2=\frac{\text{given mass}}{\text{Molar mass}}=\frac{8}{38}=0.21mol$
Finding the limiting reagent:
$ForN{H}_3=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{0.118}{2}=0.059mol\text{For}{F}_2=\frac{0.21}{5}=0.042mol$
The limiting reagent will be $F_2.$
5 moles of $F_2$ produce 1 mole of $N_2{F}_4$
$\text{0.21 moles will produce = 0.042 moles of}{N}_2{F}_4$
Amount of $N_2{F}_4$ produced $=0.042\times 104=4.368g$