Question

$Zn+H_2{SO}_4\to Zn{SO}_4+H_2$
A reaction of zinc metal with sulfuric acid produces $1.5\times {10}^{-2}mol$ of $Zn{SO}_4$ from $2.0\times {10}^{-2}mol$ of $Zn$.
What was the percent yield of this reaction?

• A. $25$%
• B. $75$%
• C. $33$%
• D. $67$%

Answer 1

The correct option is B $75$%

$1mole$ of Zn react with $1mole$ of $H_{2}S{O}_4$ produce $1mole$ of $ZnS{O}_4$.

So, to produce $1.5\times {10}^{-2}ZnS{O}_4$, $1.5\times {10}^{-2}moles$.of zinc is needed.

Here, Actual Yield $=1.5\times {10}^{-2}moles$

Theoretical Yield $=2\times {10}^{-2}moles$

$\Rightarrow$ Percent Yield $=\frac{1.5\times {10}^{-2}}{2\times {10}^{-2}}\times 100=75\%$