Question

Chlorine gas reacts with ammonia to form nitrogen gas and ammonium chloride. If 70 g of chlorine and 50 g of ammonia form 6.2 grams of nitrogen, what is the percent yield of this reaction? What is oxidized? What is reduced?

Answer 1

1)

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 70.0 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(70 g)/(70.9 g/mol)

= 0.9873 mol

Molar mass of NH3,

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

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 50.0 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(50 g)/(17.03 g/mol)

= 2.935 mol

Balanced chemical equation is:

3 Cl2 + 8 NH3 ---> N2 + 6 NH4Cl

3 mol of Cl2 reacts with 8 mol of NH3

for 0.9873 mol of Cl2, 2.633 mol of NH3 is required

But we have 2.935 mol of NH3

so, Cl2 is limiting reagent

we will use Cl2 in further calculation

Molar mass of N2 = 28.02 g/mol

According to balanced equation

mol of N2 formed = (1/3)* moles of Cl2

= (1/3)*0.9873

= 0.3291 mol

use:

mass of N2 = number of mol * molar mass

= 0.3291*28.02

= 9.221 g

% yield = actual mass*100/theoretical mass

= 6.2*100/9.221

= 67.23%

Answer: 67. %

2)

3 Cl2 + 8 NH3 —> N2 + 6 NH4Cl

Oxidation state of Cl changes from 0 to -1

So, Cl is reduced

Oxidation state of N in NH3 is -3 and N2 is 0

So, N is oxidised