Answer to Question #96124 in General Chemistry for Emily

Question

CrCl3 + AgNO3 > AgCl + Cr(NO3)3
A 0.720g sample of impure CrCl3 was dissolved in water and required 43.5 mL of 0.115 M AgNO3 solution to react with the CrCl3 in the sample. What is the mass percent of CrCl3 in the sample?

Answer 1

The reaction occurs by the following equation:

$CrCl_3 + 3AgNO_3 \rightarrow 3AgCl + Cr(NO_3)_3$

The amount of silver nitrate required for the reaction can be found:

$n(AgNO_3) = c(AgNO_3) \cdot V(AgNO_3) = 0.115 \frac{mol}{L} \cdot 0.0435 L = 0.005 mol$

According to stoichiometry of the reaction the amount of CrCl₃ is one third of silver nitrate:

$n(CrCl_3) = \frac{1}{3} \cdot n(AgNO_3)$

The mass of chromium chloride reacted with silver nitrate is:

$m(CrCl_3) = n(CrCl_3) \cdot M(CrCl_3) = \frac{1}{3} \cdot 0.005 mol \cdot 158.5 \frac{g}{mol} = 0.264 g$

The mass percent of chromium chloride in the sample is:

$\omega (CrCl_3) = \frac{m(CrCl_3)}{m(sample)} \cdot 100 \% = \frac{0.264 g}{0.720 g} \cdot 100 \% =36.7 \%$