Assuming an efficiency of 21.20 % , calculate the actual yield of magnesium nitrate formed from 143.2 g of magnesium and excess copper(II) nitrate. Mg + Cu ( NO 3 ) 2 ⟶ Mg ( NO 3 ) 2 + Cu

The actual yield of magnesium nitrate is 187.6 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of magnesium = 143.2 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

1 mole of magnesium produces 1 mole of magnesium nitrate

So, 5.967 moles of magnesium will produce = of magnesium nitrate

Now, calculating the mass of magnesium nitrate from equation 1, we get:

Molar mass of magnesium nitrate = 148.3 g/mol

Moles of magnesium nitrate = 5.967 moles

Putting values in equation 1, we get:

To calculate the actual yield of magnesium nitrate, we use the equation:

Percentage yield of magnesium nitrate = 21.20 %

Theoretical yield of magnesium nitrate = 884.906 g

Putting values in above equation, we get:

Hence, the actual yield of magnesium nitrate is 187.6 grams

particle   p is an electron and particle q is a neutron