Question

When 11.0g of Mg(s) reacts with excess water to produce 5.75g of Mg(OH)2, what is the percent yield of the reaction

Answer 1

The chemical equation is given by :

$Mg +2H_2O-> Mg(OH)_2+H_2$

From above equation :

24 gm Mg forms -----> 58 gm of $Mg(OH)_2$

So, 11 g of Mg will form = $\dfrac{11}{24}\times 58=26.58\ gm$.

Now, percentage yield is given by :

$\%\ yield = \dfrac{5.75}{26.58}\times 100\\\\\% \ yield = 21.63\%$

Therefore, the percentage yield is given by 21.63%.

Hence, this is the required solution.