Question

the interhalogen compound ClF3 reacts with NH3 to produce three products according to the figure shown here:

If 27.4 g of NH3 and 172.5 g of ClF3 are allowed to react, what masses of each product would be recovered assuming complete reaction?

____________ g N2

____________ g Cl2

____________ g HF

Answer 1

2 NH3 + 2 ClF3 —> N2 + Cl2 + 6 HF

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = 27.4 g

we have below equation to be used:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(27.4 g)/(17.034 g/mol)

= 1.609 mol

Molar mass of ClF3 = 1*MM(Cl) + 3*MM(F)

= 1*35.45 + 3*19.0

= 92.45 g/mol

mass of ClF3 = 172.5 g

we have below equation to be used:

number of mol of ClF3,

n = mass of ClF3/molar mass of ClF3

=(172.5 g)/(92.45 g/mol)

= 1.866 mol

2 mol of NH3 reacts with 2 mol of ClF3

for 1.6085 mol of NH3, 1.6085 mol of ClF3 is required

But we have 1.8659 mol of ClF3

so, NH3 is limiting reagent

we will use NH3 in further calculation

1)

Molar mass of N2 = 28.02 g/mol

From balanced chemical reaction, we see that

when 2 mol of NH3 reacts, 1 mol of N2 is formed

mol of N2 formed = (1/2)* moles of NH3

= (1/2)*1.6085

= 0.8043 mol

we have below equation to be used:

mass of N2 = number of mol * molar mass

= 0.8043*28.02

= 22.54 g

Answer: 22.5 g

2)

Molar mass of Cl2 = 70.9 g/mol

From balanced chemical reaction, we see that

when 2 mol of NH3 reacts, 1 mol of Cl2 is formed

mol of Cl2 formed = (1/2)* moles of NH3

= (1/2)*1.6085

= 0.8043 mol

we have below equation to be used:

mass of Cl2 = number of mol * molar mass

= 0.8043*70.9

= 57.02 g

Answer: 57.0 g

3)

Molar mass of HF = 1*MM(H) + 1*MM(F)

= 1*1.008 + 1*19.0

= 20.008 g/mol

From balanced chemical reaction, we see that

when 2 mol of NH3 reacts, 6 mol of HF is formed

mol of HF formed = (6/2)* moles of NH3

= (6/2)*1.6085

= 4.826 mol

we have below equation to be used:

mass of HF = number of mol * molar mass

= 4.826*20.01

= 96.55 g

Answer: 96.6 g