Question

if 7.30 g of HCl & 4.00 g of NH3 are mixed, How many grams of NH4Cl can be formed?

Answer 1

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

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = 4.0 g

we have below equation to be used:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(4.0 g)/(17.034 g/mol)

= 0.2348 mol

Molar mass of HCl = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 7.3 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(7.3 g)/(36.458 g/mol)

= 0.2002 mol

we have the Balanced chemical equation as:

NH3 + HCl ---> NH4Cl +

1 mol of NH3 reacts with 1 mol of HCl

for 0.2348 mol of NH3, 0.2348 mol of HCl is required

But we have 0.2002 mol of HCl

so, HCl is limiting reagent

we will use HCl in further calculation

Molar mass of NH4Cl = 1*MM(N) + 4*MM(H) + 1*MM(Cl)

= 1*14.01 + 4*1.008 + 1*35.45

= 53.492 g/mol

From balanced chemical reaction, we see that

when 1 mol of HCl reacts, 1 mol of NH4Cl is formed

mol of NH4Cl formed = (1/1)* moles of HCl

= (1/1)*0.2002

= 0.2002 mol

we have below equation to be used:

mass of NH4Cl = number of mol * molar mass

= 0.2002*53.49

= 10.7 g

Answer: 10.7 g