Question

Map pling Learning In the following reaction, 4514 g of lead reacts with excess oxygen forming 354.0 g of lead(I) oxide. Calculate the percent yield of the reaction. Number

Answer 1

2 Pb(s) + O2 (g) -----> 2 PbO(s)

2 moles of Pb( 2* 207 g) of lead forms 2 moles of PbO ( 2*223 g )

414 g Pb forms 446 g PbO
451.4 g Pb will form 451.4 * 446 / 414 = 486.29 g PbO ( this is theoritical yield)

But actually it formed only 354 g ( this is actual Yield)

% Yield = (Actual yield / Theoritical Yield )* 100 = (354 g / 486.29 g ) * 100 = 72.79 %

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