Question

1. If actual amount of alum obtained from 2.535 grams of aluminum sulfate is 3.5079 g , what is the percent yield of alum?

2. In a combination reaction, if 1.34794 moles of magnesium is heated with 0.2638 moles of nitrogen, to form magnesium nitride, how many grams of magnesium nitride will be produced?

If actual amount of alum obtained from 2.535 grams of aluminum sulfate is 3.5079 g , what is the percent yield of alum?

Answer 1

1) Reaction is

K2 SO4 (aq) + Al2 (SO4 )3 (aq) .................> 2 KAl(SO4 )2

one mole Al2(SO4)3 produce 2 mole alum.

2.535 gm Al2(SO4)3 = mass / molar mass = 2.535 / 342.15 = 0.00741 mole.

so theoretical yield of alum = 2 * 0.00741 * 474.3884 = 7.03 gm.

% of yield = observed yield * 100 / theoretical yield = 3.5079 * 100 / 7.03 = 49.9 %

2)

reaction is

3 Mg + N2 ..............> Mg3N2

here Mg : N2 = 3 : 1 to form one mole Mg3N2.

N2 is the limiting reactant.

grams of Mg3N2 formed = 0.2638 * 100.94 = 26.63 gm.