in the following reaction, 451.4g of lead reacts with excess oxygen forming 356.7g of lead (II) oxide. Calculate the percent yield of the reaction. 2Pb(s)+O2(g)= 2PbO(s)

Question

jfraser · Answer

First, and most important, what's the balanced reaction?

asib1214 · Answer

$$2Pb^{2+} (s) + O_2 (g) = 2Pb^{2+}O^{2-} (s)$$
451.4g                        356.7g    

106.42g/mol                223.2g/mol

1) convert mass of given substance to amount of given substance.

$$n_Pb^{2+}= 451.4g  X  1mol/106.42g $$  where grams are canceled out and you get 4.24mol

2) convert amount of given substance to amount of required substance.

$$n_Pb^{2+}O^{2-}= 4.24mol_Pb X 2mol PbO/2mol pb$$ where moles are canceled out and you get 4.24mol PbO

3) Convert amount of required substance to mass of required substance.

$$m_PbO = 4.24mol_PbO X 223.2g/1mol $$ where moles are canceled out and you get 946.368g

4) Calculate the percent yield. 

Percent yield = actual yield (Given)/ theoratical yield (Found) X 100%

                      356.7g/946.368g  X100%
                      37%
 Therefore the percent yield in the synthesis of Lead monoxide, PbO, is 37%.