A mass of 14.22g of (NH4)2SO4 (molar mass = 132.06g/mol) is dissolved in water.

Question

A mass of 14.22g of (NH4)2SO4 (molar mass = 132.06g/mol) is dissolved in water. After the solution is heated, 30.19g of Al2(SO4)3•18H2O (molar mass = 666.36g/mol) is added. calculate the theoretical yield of the resulting alum. Hint: This is a limiting reactant problem. A mass of 14.22g of (NH4)2SO4 (molar mass = 132.06g/mol) is dissolved in water. After the solution is heated, 30.19g of Al2(SO4)3•18H2O (molar mass = 666.36g/mol) is added. calculate the theoretical yield of the resulting alum. Hint: This is a limiting reactant problem.

Explanation / Answer

(NH4)2SO4 + Al2(SO4)3•18H2O --------> (NH4)2SO4.Al2(SO4)3.18H2O

no of moles of (NH4)2SO4   = w/G.M.Wt

                                          = 14.22/132.06 = 0.1076 moles

no of moles of Al2(SO4)3•18H2O = W/G.M.Wt

                                                  = 30.19/666.36   =0.0453 moles

from the balanced equation 1 mole of (NH4)2SO4 react with 1 moles of Al2(SO4)3•18H2O

limiting reagent is Al2(SO4)3•18H2O

1 mole of Al2(SO4)3•18H2O react with (NH4)2SO4 to form 1 mole of (NH4)2SO4.Al2(SO4)3.18H2O

0.0453 moles of Al2(SO4)3•18H2O react with (NH4)2SO4 to form 0.0453 mole of (NH4)2SO4.Al2(SO4)3.18H2O

mass of (NH4)2SO4.Al2(SO4)3.18H2O = no of moles * molar mass

                                                          = 0.0453*798.42 = 36.16 gm

the theoretical yield of the resulting alum is 36.16 gm

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