Question

Assuming an efficiency of 47.30 % , calculate the actual yield of magnesium nitrate formed from 119.3 g of magnesium and excess copper(II) nitrate.

Mg + Cu ( NO 3 ) 2 ⟶ Mg ( NO 3 ) 2 + Cu

Answer 1

Actual yield of
`Mg(NO_(3))_(2)` is 344.3 g

Step-by-step explanation:

According to balanced equation, 1 mol of Mg produces 1 mol of
`Mg(NO_(3))_(2)`

Molar mass of Mg = 24.305 g/mol and molar mass of
`Mg(NO_(3))_(2)` = 148.3 g/mol

So, 24.305 of Mg produces 148.3 g of
`Mg(NO_(3))_(2)`

Hence, 119.3 of Mg produces
`(148.3* 119.3)/(24.305)g` of
`Mg(NO_(3))_(2)` or 727.9 g of
`Mg(NO_(3))_(2)`

Hence, theoretical yield of
`Mg(NO_(3))_(2)` = 727.9 g

So, actual yield of
`Mg(NO_(3))_(2)` =
`(727.9* 0.4730)g` = 344.3 g