Question

In the reaction 2Pb(s)+O2(g)⟶2PbO(s) 451.4 g of lead reacts with excess oxygen, forming 301.8 g of...

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.

Homework Answers

Answer #1

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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