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

Respuesta :

Answer:

Actual yield of [tex]Mg(NO_{3})_{2}[/tex] is 344.3 g

Explanation:

According to balanced equation, 1 mol of Mg produces 1 mol of [tex]Mg(NO_{3})_{2}[/tex]

Molar mass of Mg = 24.305 g/mol and molar mass of [tex]Mg(NO_{3})_{2}[/tex] = 148.3 g/mol

So, 24.305 of Mg produces 148.3 g of [tex]Mg(NO_{3})_{2}[/tex]

Hence, 119.3 of Mg produces [tex]\frac{148.3\times 119.3}{24.305}g[/tex] of [tex]Mg(NO_{3})_{2}[/tex] or 727.9 g of [tex]Mg(NO_{3})_{2}[/tex]

Hence, theoretical yield of [tex]Mg(NO_{3})_{2}[/tex] = 727.9 g

So, actual yield of [tex]Mg(NO_{3})_{2}[/tex] = [tex](727.9\times 0.4730)g[/tex] = 344.3 g

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