Question

The reaction shown below is used to destroy Freon-12 (CF₂Cl₂), preventing its release into the atmosphere. What mass of NaF will be formed if 240 kg of CF₂Cl₂ and 480 kg of Na₂C₂O₄ are heated and allowed to react to completion?

CF₂Cl₂ + 2 Na₂C₂O₄ → 2 NaF + 2 NaCl + C + 4 CO₂

Answer 1

Molar mass of CF2Cl2,

MM = 1*MM(C) + 2*MM(F) + 2*MM(Cl)

= 1*12.01 + 2*19.0 + 2*35.45

= 120.91 g/mol

mass(CF2Cl2)= 240 kg = 240000 g

number of mol of CF2Cl2,

n = mass of CF2Cl2/molar mass of CF2Cl2

=(240000 g)/(120.91 g/mol)

= 1.985*10^3 mol

Molar mass of Na2C2O4,

MM = 2*MM(Na) + 2*MM(C) + 4*MM(O)

= 2*22.99 + 2*12.01 + 4*16.0

= 134 g/mol

mass(Na2C2O4)= 480 kg = 480000 g

number of mol of Na2C2O4,

n = mass of Na2C2O4/molar mass of Na2C2O4

=(480000 g)/(134 g/mol)

= 3.582*10^3 mol

1 mol of CF2Cl2 reacts with 2 mol of Na2C2O4

for 1984.9475 mol of CF2Cl2, 3969.895 mol of Na2C2O4 is required

But we have 3582.0896 mol of Na2C2O4

so, Na2C2O4 is limiting reagent

we will use Na2C2O4 in further calculation

Molar mass of NaF,

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

= 1*22.99 + 1*19.0

= 41.99 g/mol

According to balanced equation

mol of NaF formed = (2/2)* moles of Na2C2O4

= (2/2)*3582.0896

= 3582.0896 mol

mass of NaF = number of mol * molar mass

= 3.582*10^3*41.99

= 1.504*10^5 g

= 150 Kg

Answer: 150 Kg