Question

3. Aluminum chloride reacts with sulfuric acid to form aluminum sulfate and hydrochloric acid according to the balanced chemical equation below. 2A1C13(aq) + 3H2SO4(aq) → Al2(SO4)3(aq) + 6HCl(aq) 133.340 g/mol 98.079 g/mol 342.150 g/mol 36.460 g/mol Molar Masses: a) How many grams of sulfuric acid react with 25.0 grams of aluminum chloride? b) What is the theoretical yield of HCl if sulfuric acid is in excess?

Answer 1

a)

Molar mass of AlCl3,

MM = 1*MM(Al) + 3*MM(Cl)

= 1*26.98 + 3*35.45

= 133.33 g/mol

mass of AlCl3 = 25 g

mol of AlCl3 = (mass)/(molar mass)

= 25/1.333*10^2

= 0.1875 mol

According to balanced equation

mol of H2SO4 formed = (3/2)* moles of AlCl3

= (3/2)*0.1875

= 0.2813 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass of H2SO4 = number of mol * molar mass

= 0.2813*98.09

= 27.59 g

Answer: 27.6 g

b)

Molar mass of AlCl3,

MM = 1*MM(Al) + 3*MM(Cl)

= 1*26.98 + 3*35.45

= 133.33 g/mol

mass of AlCl3 = 25 g

mol of AlCl3 = (mass)/(molar mass)

= 25/1.333*10^2

= 0.1875 mol

According to balanced equation

mol of HCl formed = (6/2)* moles of AlCl3

= (6/2)*0.1875

= 0.5625 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = number of mol * molar mass

= 0.5625*36.46

= 20.51 g

Answer: 20.5 g