Question

Consider the reaction.2 Pb(s) + O₂(g) →→→ 2 PbO(s)->An excess of oxygen reacts with 451.4 g of lead, forming 316.9 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

In this question, we have the following reaction:

2 Pb + O2 -> 2 PbO

We have to find the percent yield of this reaction, we have:

451.4 grams of Pb

316.9 grams of PbO, this is the actual yield, we will have to find the theoretical yield now

In order to find the theoretical yield, we need to know how many moles of Pb we have in 451.4 grams, the molar mass of Pb is 207.2g/mol

207.2g = 1 mol

451.4 g = x moles

207.2x = 451.4

x = 451.4/207.2

x = 2.18 moles of Pb in 451.4 grams

According to the molar ratio of Pb and PbO, we have the ratio of 2 moles of Pb to form 2 moles of PbO, therefore if we have 2.18 moles of Pb, we will also have 2.18 moles of PbO.

Now we need to find the theoretical mass of PbO, using its molar mass, 223.2g/mol

223.2g = 1 mol

x grams = 2.18 moles

x = 2.18 * 223.2

x = 486.6 grams of PbO is the theoretical yield

Now to find the percent yield, we need to use the following formula:

%yield = (actual yield/theoretical yield)*100

%yield = (316.9/486.6)*100

%yield = 65

The percent yield of this reaction will be 65%