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

Step-by-step 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}=(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 =
`(1)/(1)* 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* 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}}* 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}* 100\\\\\text{Actual yield of magnesium nitrate}=(21.20* 884.906)/(100)=187.6g`

Hence, the actual yield of magnesium nitrate is 187.6 grams