Question

Consider the reaction. 2 Pb(s) + 0,(g) 2 PbO(s) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of lead(II) oxide. Calculate the percent yield of the reaction. percent yield:

An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of Lead (II) oxide

calculate the percent yield of the reaction

Answer 1

Molar mass of Pb = 207.2 g/mol

mass of Pb = 4.514*10^2 g

mol of Pb = (mass)/(molar mass)

= 4.514*10^2/2.072*10^2

= 2.179 mol

According to balanced equation

mol of PbO formed = (2/1)* moles of Pb

= (2/1)*2.179

= 4.357 mol

Molar mass of PbO,

MM = 1*MM(Pb) + 1*MM(O)

= 1*207.2 + 1*16.0

= 223.2 g/mol

mass of PbO = number of mol * molar mass

= 4.357*2.232*10^2

= 9.725*10^2 g

% yield = actual mass*100/theoretical mass

= 3.675*10^2*100/9.725*10^2

= 37.79 %

Answer: 37.79 %