college chemistry 1

Question

An excess of oxygen reacts with 451.4 g of lead, forming 347.0 of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

First, write out the balanced chemical equation:

O₂ + 2Pb --> 2PbO

Then, calculate the theoretical yield of lead(II) oxide by using the 451.4g of lead provided in the question.

Molar mass of lead (Pb) = 207.2g/mol

Molar mass of lead(II) oxide (PbO) = 207.2g/mol + 16.00g/mol = 223.2 g/mol

1. Convert g of lead to mols of lead using molar mass of lead from periodic table
2. Convert mols of lead to mols of lead(II) oxide using the coefficients from the balanced equation
3. Convert mols of lead(II) oxide to g of lead(II) oxide using molar mass from periodic table

451.4g Pb x 1 mol Pb x 2 mol PbO x 223.2g PbO = 486.26g PbO

207.2g Pb. 2 mol Pb. 1 mol PbO

Next, plug the theoretical yield (solved for above) and actual yield (provided in question) into the equation for percent yield.

% yield = actual yield x 100

theoretical yield

= 347.0g x 100

486.26g

= 71.36%