Question

451.4 g of lead reacts with excess oxygen forming 377.0 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

You will need to know this equation:

mol= mass/MW

MW= molecular weight, taken from the periodic table.

In order to find the percent yield we need to find the theoretical yield first. This means, the amount of product that could be produced with the amount of reagents given. So, first calculate moles of Pb:

Pb + 1/2O2 ----> PbO

mol Pb= 451.4g/207.2g/mol= 2.18 mol

Now, looking at the balanced reaction we can see that 1 mol of Pb produces 1 mol of PbO. This means that if we have 2.18 mol of Pb, theoretically the reaction will produce 2.18 mol of product. Now transform this amount of product to mass:

mass of PbO theoretically produced= 2.18mol x 223.2g/mol = 486.3g (this is the theoretical yield).

Now the percent yield:

%yield= amount obtained from the reaction/theoretical yield x 100

%yield= 313.1g/486.3g x 100= 64.4%

Check the significant figures, the % is most of the times written with 2 significant figures, so it will be 64%.

Let me know if you have any doubt!