Question

$Zn+{{H}_{2}}S{{O}_{4}}\to ZnS{{O}_{4}}+{{H}_{2}}$
 A reaction of zinc metal with sulfuric acid produces $1.5\times {{10}^{-3}}mol$ of $ZnS{{O}_{4}}$ 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

Hint: By looking at the stoichiometry of the given reaction, we would be able to deduce the experimental and theoretical yield. On dividing the experimental yield by theoretical yield and multiplication with 100 will give us the percent yield for this reaction.

Complete step by step answer:
- In the given reaction zinc metal is reacted with sulfuric acid to form zinc sulphate and hydrogen gas. This reaction is a general example of a reaction between a metal and acid producing the metal salt and hydrogen gas.
-We are asked to find the percent yield of the reaction. So, let's see what's the idea behind percent yield. It is the ratio between what is experimentally obtained and what is theoretically calculated and this is multiplied by 100 in order to get the percentage value. Thus, the equation for calculating percent yield can be written as follows
$Percent\text{ }yield=\dfrac{Actual\text{ }yield}{Theoretical~yield}\times 100$

- When we look at the balanced chemical reaction between zinc and sulfuric acid, we could understand that the mole ratio between reactants and products is 1:1.
-This implies that if we are taking one mole of reactant, then ideally or theoretically speaking there should be one mole of the product. But this doesn’t happen and here comes the relevance of percent yield.

 -In the reaction ,it's given that only $1.5\times {{10}^{-3}}mol$ of $ZnS{{O}_{4}}$ is formed from $2.0\times {{10}^{-2}}mol$ of Zn. Thus, here the actual yield we have obtained is $1.5\times {{10}^{-3}}mol$and the theoretical yield we were expecting is $2.0\times {{10}^{-2}}mol$. On substituting these values in the above equation of percent yield we get,
 Percent yield = $\dfrac{1.5\times {{10}^{-3}}}{2.0\times {{10}^{-2}}}\times 100$
 = 75 $%$
So, the correct answer is “Option B”.

Note: It should be noted that if we get the actual value as same as the theoretical yield, it implies 100 $%$ yield. Normally we obtain a value less than 100 $%$.As per the percent yield, it's impossible to obtain a value above 100 $%$ and if this happens, then there will be some experimental or calculation error.