Question

A reaction mixture contains 4.00 g of ammonia, NH3 and 14.0 g of fluorine, F2. What is the theoretical yield of N2F4 for the following reaction? 2 NH3 + 5 F2 yields N2F4 + 6 HF

Answer 1

Solution :-

Balanced reaction equation

2 NH3 + 5 F2 ------ > N2F4 + 6 HF

Lets calculate the mass of the N2F4 produced from the mole ratio of the each reactant

calculating mass of N2F4 produced from the 4.00 g NH3

(4.00 g NH3*1 mol / 17.03 g) *(1 mol N2F4 / 2 mol NH3) * (104.01 g / 1 mol N2F4) = 12.2 g N2F4

No calculating from the 14.0 g F2

(14.0 g F2 * 1 mol / 37.996 g)*(1 mol N2F4 / 5 mol F2)*(104.01 g / 1 mol N2F4) = 7.66 g N2F4

The mass of the N2F4 produced from the F2 is less therefore the F2 is the limiting reactant and hence maximum mass of N2F4 that can be produed = 7.66 g N2F4

So the theoretical yield of the N2F4 = 7.66 g