Question

A 12.00g sample of MgCl2 was dissolved in water. The solution was then treated with 0.2500mol of AgNO3 to precipitate all the chloride ions from the solution. Calculate the purity (as a mass percentage) of MgCl2 in the sample?

Answer 1

Answer: The percentage purity of magnesium chloride in the sample is 96.04 %

Explanation:

The chemical equation for the reaction of silver nitrate and magnesium chloride follows:

$MgCl_2+2AgNO_3\rightarrow 2AgCl+Mg(NO_3)_2$

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of magnesium chloride

So, 0.2500 moles of silver nitrate will react with = $\frac{1}{2}\times 0.2500=0.125mol$ of magnesium chloride

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of magnesium chloride = 0.125 moles

Molar mass of magnesium chloride = 92.2 g/mol

Putting values in above equation, we get:

$0.125mol=\frac{\text{Mass of magnesium chloride}}{92.2g/mol}\\\\\text{Mass of magnesium chloride}=(0.125mol\times 92.2g/mol)=11.525g$

To calculate the percentage purity of sample, we use the equation:

$\%\text{ purity of sample}=\frac{\text{Mass of pure sample}}{\text{Mass of impure sample}}\times 100$

Mass of pure magnesium chloride = 11.525 grams

Mass of impure magnesium chloride = 12.00 grams

Putting values in above equation, we get:

$\%\text{ purity of magnesium chloride}=\frac{11.525g}{12.00g}\times 100\\\\\%\text{ purity of magnesium chloride}=96.04\%$

Hence, the percentage purity of magnesium chloride in the sample is 96.04 %