Answer:
[tex]\% m=40.46\%[/tex]
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:
[tex]BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl[/tex]
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:
[tex]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[/tex]
Thus, the mass percentage is calculated as shown below:
[tex]\% m=\frac{1.479g}{3.656g}*100 \% \\\\\% m=40.46\%[/tex]
Regards!
