Question

1. In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.8 g of lead(II) oxide. Calculate the percent yield of the reaction.

2Pb(s)+O2(g) = 2PbO(s)

2. Assuming an efficiency of 32.60%, calculate the actual yield of magnesium nitrate formed from 119.8 g of magnesium and excess copper(II) nitrate.

Mg + Cu(NO3)2 = Mg(NO3)2 + Cu

3. Combining 0.282 mol of Fe2O3 with excess carbon produced 18.8 g of Fe.

Fe2O3 + 3C = 2Fe + 3CO

What is the actual yield of iron in moles?

What was the theoretical yield of iron in moles?

What was the percent yield?

Answer 1

1. From balanced equation it is clear that 2 mole of Pb gives 2 mole of PbO

Theoretical mass of PbO calculated is 486 gram and experimental mas of PbO is 365.8 gram

Hence % yield = experimental yield/theoretical yield ×100=75.2

2. From balanced equation it is clear that moles of Mg is equal to moles of Mg(NO3)2

Moles of Mg = 5 = moles of Mg(NO3)2

mass of Mg(NO3)2 = 5 × 148 = 738 gram

But efficiency is 32.60 hence actual yield = 738×32.60/100=240.83 gram

3. Actual yield of iron in mole = 0.335

Theoretical yield of iron in mole = 0.564

% yield of iron = 18.8/31.58×100= 59.53