Question

11.26 Determine the theoretical yield of HF, in grams, when 49.8 g of NH3 reacts with 49.8 g of F2. What is the percent yield if 7.22 g of HF is actually formed during the reaction? 2NH3(g) + 5F2(8) N2F4(8) + 6HF(8)

Answer 1

1)

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 49.8 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(49.8 g)/(17.03 g/mol)

= 2.924 mol

Molar mass of F2 = 38 g/mol

mass(F2)= 49.8 g

use:

number of mol of F2,

n = mass of F2/molar mass of F2

=(49.8 g)/(38 g/mol)

= 1.311 mol

Balanced chemical equation is:

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

2 mol of NH3 reacts with 5 mol of F2

for 2.924 mol of NH3, 7.309 mol of F2 is required

But we have 1.311 mol of F2

so, F2 is limiting reagent

we will use F2 in further calculation

Molar mass of HF,

MM = 1*MM(H) + 1*MM(F)

= 1*1.008 + 1*19.0

= 20.008 g/mol

According to balanced equation

mol of HF formed = (6/5)* moles of F2

= (6/5)*1.311

= 1.573 mol

use:

mass of HF = number of mol * molar mass

= 1.573*20.01

= 31.47 g

Answer: 31.5 g

2)

% yield = actual mass*100/theoretical mass

= 7.22*100/31.47

= 23.0 %

Answer: 23.0 %