In the following reaction, 451.4 grams of lead reacts with excess oxygen forming 359.9 grams of lead(II) oxide. Calculate the percent yield of the reaction? 2Pb(s)+O2(g) --> 2PbO(s)

Question

39937 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-28T23:11:45

Consider,

#2Pb(s) + O_2(g) to 2PbO(s)#

Let's first calculate the theoretical yield of the oxide given that mass of lead metal.

#451.4"g" * "mol"/(207.2"g") * (2PbO)/(2Pb) * (223.2"g")/(PbO) approx 486.3"g"# of lead oxide

should be theoretically produced in this reaction.

The actual yield was less than this, though.

#(359.9"g")/(486.3"g") * 100% approx 74.01%#

is the percent yield of this reaction. Not bad!