Question

7. In the Thermit reaction, 2Al(s) + Cr2O; (s) — 2Cr (s) + A1:0(s) a) What is the mass of chromium(III) oxide needed to produce 150 kg of chromium? b) What will be the corresponding mass of Al O, formed in (a)? c) Calculate the percentage yield when 180 g of chromium are obtained from a reaction between 100 g of aluminum and 400 g of chromium(III) oxide.

Answer 1

a) 150 kg of Cr = 150*10^3 g = 150*10^3 g / 51.996g/mol = 2.885*10^3 mol

As per the balanced equation, one mole of Cr2O3 gives 2 mole of Cr.

Hence to have 2.885*10^5 mol of Cr we need = (2.885*10^3)/2 = 1.443*10^3 mol of Cr2O3

Mass of 1.443*10^3 mol of Cr2O3 = (1.443*10^3)*molar mass = (1.443*10^3 )* 151.99 = 219.3 kg

Thus to produce 150 kg of Cr we need 219.3 kg of Cr2O3.

b) 1.443*10^3 mol of Cr2O3 reacts with =2* 1.443*10^3 mol = 2.886*10^3 mole of Al

mass of Al = 2.886*10^3 mol * 26.98g/mol = 77.864*10^3 g = 77.864 kg of Al

2.886*10^3 mol of Al gives = (2.886*10^3 mol )/2 = 1443 mol of Al2O3

Mass of Al2O3 = 1443 mol * 101.96g/mol = 147.13 kg

c) Mole of 400 g Cr2O3 = 400g/151.99g/mol = 2.6318 mol of Cr2O3

moles of 100 g Al = 100g/26.98g/mol = 3.7065 mol of Al

Thus the Al is the limiting reagent and Cr2O3 is in excess.

3.7065 mol of Al gives = 3.7065 mol of Cr

Theoretical yield of Cr = 3.7065 mol * 51.996g/mol = 192.7 g of Cr

Practical yieldis given = 180 g

Hence, % yield = (practical/theoretical yield)*100 = (180/192.7)*100 = 93.41%