Question

A 12.00g sample of MgCl2 was dissolved in water. 0.2500mol of AgNO3 was required to precipitate all the chloride ions from the solution. Calculate the purity (as a mass percentage) of MgCl2 in the sample. Your answer should have four significant figures (round to the nearest hundredth of a percent).

Answer 1

Answer:

$Purity=99\%$

Explanation:

Hello,

In this case, the undergoing precipitation reaction is:

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

Thus, for the 0.2500 moles of silver nitrate, the following mass of magnesium chloride is consumed (consider their 2:1 molar ratio):

$m_{MgCl_2}=0.2500molAgNO_3*\frac{1molMgCl_2}{2molAgNO_3} *\frac{95.2gMgCl_2}{1molMgCl_2} \\\\m_{MgCl_2}=11.90gMgCl_2$

Therefore, the purity of the sample is:

$Purity=\frac{11.90g}{12.00g}*100\%\\ \\Purity=99\%$

Best regards.

Answer 2

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%