Question

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

Answer 1

Answer: The actual yield of magnesium nitrate is 187.6 grams

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of magnesium = 143.2 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

$\text{Moles of magnesium}=\frac{143.2g}{24g/mol}=5.967mol$

For the given chemical equation:

$Mg+Cu(NO_3)_2\rightarrow Mg(NO_3)_2+Cu$

By Stoichiometry of the reaction:

1 mole of magnesium produces 1 mole of magnesium nitrate

So, 5.967 moles of magnesium will produce = $\frac{1}{1}\times 5.967=5.967mol$ 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:

$5.967mol=\frac{\text{Mass of magnesium nitrate}}{148.3g/mol}\\\\\text{Mass of magnesium nitrate}=(5.967mol\times 148.3g/mol)=884.906g$

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

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of magnesium nitrate = 21.20 %

Theoretical yield of magnesium nitrate = 884.906 g

Putting values in above equation, we get:

$21.20=\frac{\text{Actual yield of magnesium nitrate}}{884.906g}\times 100\\\\\text{Actual yield of magnesium nitrate}=\frac{21.20\times 884.906}{100}=187.6g$

Hence, the actual yield of magnesium nitrate is 187.6 grams