Question

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 321.9 g of lead(II) oxide. Calculate the percent yield of the reaction. 2Pb(s)+O2(g)-->2PbO(s)

Answer 1

Molar mass of Pb is 207.2 g/mol

Molar mass of PbO = At.mass of Pb + At.mass of O

                             = 207.2 + 16.0

                             = 223.2 g/mol

2Pb(s)+O₂(g) 2PbO(s)

According to the balanced equation,

2 moles of Pb reacts with excess of oxygen produces 2 moles of PbO

2 x207.2 g of Pb reacts with excess of oxygen produces 2x223.2 g of PbO

451.4 g of Pb reacts with excess of oxygen produces M g of PbO

M = ( 2x223.2x451.4) / (2x207.2)

   = 486.3 g of PbO    ----> This is the theoretical yield

Given actual yield of the reaction is 321.9 g

% yield = ( actual yield / Theoretical yield) x 100

           = ( 321.9 / 486.3) x 100

           = 66.2 %

Therefore the percent yield is 66.2%