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: 99. 17%

Step-by-step explanation:

MgCl2(aq)+2AgNO3(aq)⟶2AgCl(s)+Mg(NO3)2(aq)

(0.2500 mol AgNO3 × 1 mol (MgCl2) /2 mol (AgNO3) × 95.211 g MgCl2 /1 mol MgCl2)

divided by 12.00 g sample = 0.99178 X 100 ≈ 99.18%

Answer 2

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

Step-by-step 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 =
`(1)/(2)* 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* 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}}* 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}=(11.525g)/(12.00g)* 100\\\\\%\text{ purity of magnesium chloride}=96.04\%`

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