Question 7.CC.10: If 22.6 g of Al reacts completely with O2, and 37.8 g of Al2......
If 22.6 g of Al reacts completely with O_2, and 37.8 g of Al_2O_3 is obtained, what is the percent yield of Al_2O_3 for the reaction?
4 Al (s)+3 O _2( g ) \longrightarrow 2 Al _2 O _3(s)Step-by-Step
Calculation of theoretical yield:
\underset{\text{Three SFs}}{22.6\cancel{ g\ Al}} \times \underset{\text{Four SFs}}{\overset{\text{Exact}}{\frac{1\cancel{ \text { mole } Al} }{26.98\cancel{ g\ Al} }}} \times \underset{\text{Exact}}{\overset{\text{Exact}}{\frac{2 \cancel{\text { moles } Al_2O _3}}{4\cancel{ \text { moles }Al} }}} \times \underset{\text{Exact}}{\overset{\text{Five SFs}}{\frac{101.96\ g\ Al_2O _3}{1 \cancel{\text { mole }Al_2O_3}}}}\\=\underset{\text{Three SFs}}{42.7\ g\text{ of }Al_2O _3 }\quad\text{Theoretical yield}Calculation of percent yield:
\frac{\text { actual yield (given) }}{\text { theoretical yield (calculated) }} \times 100 \%
=\underset{\text{Three SFs}}{\overset{\text{Three SFs}}{\frac{37.8\ \cancel{g Al _2 O _3}}{42.7\ \cancel{g { Al }_2 O _3}}}} \times 100 \%
=\underset{\text{Three SFs}}{88.5 \%}
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