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

Answer:

Actual yield of $Mg(NO_{3})_{2}$ is 344.3 g

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 $\frac{148.3\times 119.3}{24.305}g$ 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 $Mg(NO_{3})_{2}$ = $(727.9\times 0.4730)g$ = 344.3 g