Question

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 381.4 g of lead(II) oxide. Calculate the percent yield of the reaction.

2Pb(s) + O2(g) ----> 2PbO(s)

Answer 1

moles of Pb = mass of Pb / molar mass of Pb = 451.4 g / 207.2 g/mol = 2.178 mol

from the balanced equation it is very clear that

2 mol of Pb will give 2 mol of PbO or in other words

1 mol of Pb will give 1 mol of PbO accordingly

2.178 mol will produce 2.178 mole of PbO

so theritical moles of PbO = 2.178

theritical mass of PbO = moles x molar mass = 2.178 mol x 223.2 g/mol = 486.25 g

% of yield = [actual yield / theritical yield] x 100

= [ 381.4 g / 486.25 g] x 100

= 78.43%