Question

A mixture of BaCl2 and NaCl is analyzed by precipitating all the barium as BaSO4. After addition of an excess of Na2SO4 to a 3.656-g sample of the mixture, the mass of precipitate collected is 1.658 g. What is the mass percentage of barium chloride in the mixture

Answer 1

Answer:

$\% m=40.46\%$

Explanation:

Hello there!

in this case, according to the given information, it turns out firstly necessary for us to write up the chemical equation as shown below:

$BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl$

Thus, we calculate the mass of BaCl2 stoichiometrically related to the produced 1.658 g of precipitate in order to discard it from the sample:

$m_{BaCl_2}=1.658gBaSO_4*\frac{1molBaSO_4}{233.38 gBaSO_4} *\frac{1molBaCl_2}{1molBaSO_4}*\frac{208.23 gBaCl_2}{1molBaCl_2}\\\\m_{BaCl_2}=1.479gBaCl_2$

Thus, the mass percentage is calculated as shown below:

$\% m=\frac{1.479g}{3.656g}*100 \% \\\\\% m=40.46\%$

Regards!