Answer :
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:
[tex]MgCl_2+2AgNO_3\rightarrow 2AgCl+Mg(NO_3)_2[/tex]
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 = [tex]\frac{1}{2}\times 0.2500=0.125mol[/tex] of magnesium chloride
To calculate the mass for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of magnesium chloride = 0.125 moles
Molar mass of magnesium chloride = 92.2 g/mol
Putting values in above equation, we get:
[tex]0.125mol=\frac{\text{Mass of magnesium chloride}}{92.2g/mol}\\\\\text{Mass of magnesium chloride}=(0.125mol\times 92.2g/mol)=11.525g[/tex]
To calculate the percentage purity of sample, we use the equation:
[tex]\%\text{ purity of sample}=\frac{\text{Mass of pure sample}}{\text{Mass of impure sample}}\times 100[/tex]
Mass of pure magnesium chloride = 11.525 grams
Mass of impure magnesium chloride = 12.00 grams
Putting values in above equation, we get:
[tex]\%\text{ purity of magnesium chloride}=\frac{11.525g}{12.00g}\times 100\\\\\%\text{ purity of magnesium chloride}=96.04\%[/tex]
Hence, the percentage purity of magnesium chloride in the sample is 96.04 %
Answer: 99. 17%
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%