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 =
`(11)/(24)* 58=26.58\ gm`.

Now, percentage yield is given by :

`\%\ yield = (5.75)/(26.58)* 100\\\\\% \ yield = 21.63\%`

Therefore, the percentage yield is given by 21.63%.

Hence, this is the required solution.