In the reaction
2Pb(s)+O2(g)⟶2PbO(s)
451.4 g of lead reacts with excess oxygen, forming 301.8 g of lead(II) oxide. Calculate the percent yield of the reaction.
Molar mass of Pb = 207.2 g/mol
mass of Pb = 451.4 g
mol of Pb = (mass)/(molar mass)
= 451.4/207.2
= 2.1786 mol
From balanced chemical reaction, we see that
when 2 mol of Pb reacts, 2 mol of PbO is formed
mol of PbO formed = moles of Pb
= 2.1786 mol
Molar mass of PbO = 1*MM(Pb) + 1*MM(O)
= 1*207.2 + 1*16.0
= 223.2 g/mol
mass of PbO = number of mol * molar mass
= 2.1786*223.2
= 486.2571 g
% yield = actual mass*100/theoretical mass
= 301.8*100/486.2571
= 62.1 %
Answer: 62.1 %
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