Question

How much 0.15 M Na2SO4= solution is needed to precipitate the BaSO4 in which AgNO3 contains 1 g per 20 ml are needed to precipitate the AgCl from a 0.5000 g sample of BaCl2∙2H2O?

Answer 1

mass percent of $BaCl_2$ in $BaCl_2*2H_2O$ is

$mass \% = \frac{M(BaCl_2)}{M(BaCl_2*2H_2O)}= \frac{208.23}{244.27}= 0.85$

$m(BaCl_2)= mass \%\times m(BaCl_2*2H_2O)= 0.85*0.5000 = 0.425 g$

a) $Na_2SO_4+ BaCl_2 = BaSO_4+ 2NaCl$

$moles (Na_2SO_4)=0.425 g BaCl_2 (\frac{1 mol BaCl_2}{208.23 g BaCl_2})(\frac{1 mol Na_2SO_4}{1 mole BaCl_2})= 0.002041 mol Na_2SO_4$

$V=\frac{n}{c}=\frac{0.002041}{0.15} = 0.01361 L = 136.1 mL$

b) $2AgNO_3 + BaCl_2 = Ba(NO_3)_2 +2AgCl$

$mass (AgNO_3) = 0.425 g BaCl_2(\frac{1 mole BaCl_2}{208.23 g BaCl_2})(\frac{2 mole AgNO_3}{1 mole BaCl_2})(\frac {169.87 g AgNO_3}{1 mole AgNO_3})= 0.6934 g AgNO_3$

$V_{solution}= 0.6934 g (\frac{20 mL}{1 g})=13.88 mL$