Question

A student obtains a 10.0g sample of a white powder labeled as BaCl2. After completely dissolving the powder in 50.0mL of distilled water, the student adds excess Na2SO4(s), which causes a precipitate of BaSO4(s) to form, as represented by the equation above. The student filters the BaSO4(s), rinses it, and dries it until its mass is constant. Which of the following scientific questions could best be answered based on the results of the experiment?

Answer 1

Answer:

Whether barium chloride solution was pure

Explanation:

We may answer whether barium chloride was pure. The sequence of this experiment might be depicted by the following balanced chemical equations:

$BaCl_2 (aq)\rightarrow Ba^{2+} (aq) + 2 Cl^- (aq)$

$Ba^{2+} (aq) + SO_4^{2-} (aq)\rightarrow BaSO_4 (s)$

Having a total sample of 10.0 grams, we would firstly find the mass percentage of barium in barium chloride:

$\omega_{Ba^{2+}} = \frac{M_{Ba}}{M_{BaCl_2}} = \frac{137.327 g/mol}{208.23 g/mol\cdot 100\% = 65.95 \%$

This means in 10.0 g, we have a total of:

$m_{Ba^{2+}} = 10.0 g\cdot 0.6595 = 6.595 g$ of barium cations.

The precipitate is then formed and we measure its mass. Having its mass determined, we'll firstly find the percentage of barium in barium sulfate using the same approach:

$\omega_{Ba} = \frac{137.327 g/mol}{233.38 g/mol}\cdot 100\% = 58.84 \%$

Multiplying the mass we obtained by the fraction of barium will yield mass of barium in barium sulfate. Then:

• if this number is equal to 6.595 g, we have a pure sample of barium chloride;
• if this number is lower than 6.595 g, this means we have an impure sample of barium chloride, as we were only able to precipitate a fraction of 6.595 g.