Assuming an efficiency of 33.50%, calculate the actual yield of magnesium nitrate formed from 135.5 g of magnesium and excess copper(I) nitrate. Mg + Cu(NOz)2 • Mg(NOg)2 + Cu

Answer 1

Hi Catlyn,

We'll need to balance the equation first to get the correct number of moles for each reactant and product:

Mg + 2 CuNO₃ → Mg(NO₃)₂ + 2 Cu

Since the balanced equation gives mole ratios, it's best to convert the given mass to moles so that we can use the equation. To do this conversion, we need the molar mass of Mg (from periodic table)

1 mol Mg = 24.305 g Mg

We'll divide by the molar mass: 135.5 g × 1 mol/24.305 g = 5.575 mols Mg

The balanced equation shows the relationship between Mg and Mg(NO₃)₂

1 mole Mg: 1 mole Mg(NO₃)₂

Since it's a 1:1 ratio, we will get an equal no. of moles of Mg(NO₃)₂

This means that the number of moles of magnesium nitrate produced will be 5.575 mols

We convert this to grams by using the molar mass of Mg(NO₃)₂ which is 148.30 g/mol

Multiply mols by molar mass to get grams: 5.575 mols × 148.30 g/mol = 826.8 g

That would be the mass we expect if the reaction were 100% efficient. However, the question says the reaction is 33.50% efficient so we have to take this into consideration.

(33.50/100) × 826.8 g = 277.0 g Mg(NO₃)₂