Question

Aluminum metal reacts with sulfuric acid according to the following equation: 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(s) + 3H2(g) If 12.9 g of aluminum reacts with excess sulfuric acid, and 62.4 g of Al2(SO4)3 are collected, what is the percent yield of Al2(SO4)3?

Answer 1

Molar mass of Al = 26.98 g/mol

mass of Al = 12.9 g

mol of Al = (mass)/(molar mass)

= 12.9/26.98

= 0.4781 mol

Balanced chemical equation is:

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(s) + 3H2(g)

According to balanced equation

mol of Al2(SO4)3 formed = (1/2)* moles of Al

= (1/2)*0.4781

= 0.2391 mol

Molar mass of Al2(SO4)3,

MM = 2*MM(Al) + 3*MM(S) + 12*MM(O)

= 2*26.98 + 3*32.07 + 12*16.0

= 342.17 g/mol

mass of Al2(SO4)3 = number of mol * molar mass

= 0.2391*3.422*10^2

= 81.8 g

% yield = actual mass*100/theoretical mass

= 62.4*100/81.8

= 76.28 %

Answer: 76.3 %