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Concentration Terms - An Introduction (w/w etc)

50 mL of 20.8...

Question

50 mL of 20.8% (w/v) $B a C l_{2} (a q)$ and 100 mL of 9.8% (w/v) $H_{2} S O_{4}$ solutions are mixed. Find the mass the $B a S O_{4}$ formed.
(Molar mass of Ba = 137 g/mL)

A

23.32 g

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B

2.06 g

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C

0.126 g

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D

11.65 g

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Solution

The correct option is D 11.65 g
$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$
Mass of $B a C l_{2} = \frac{20.8}{100} \times 50 m L$
=10.4 g
moles of $B a C l_{2} = \frac{10.4}{208} = 0.05 m o l$
Mass of $H_{2} S O_{4} = \frac{9.8}{100} \times 100 = 9.8 g$
Moles of $H_{2} S O_{4} = \frac{9.8}{98} = 0.1 m o l$
Therefore, $B a C l_{2}$ is the limiting reagent.
Moles of $B a S O_{4}$ formed is 0.05 mol
Mass of $B a S O_{4}$ formed = 0.05 x 233 g
= 11.65 g
Therefore, the mass of $B a S O_{4}$ formed is 11.65 g.

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Similar questions

Q. 50 ml of 20.8% (w/V)

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Q. 100 mL of 20.8%

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B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

50 m l

B a C l_{2}

(20.8 % w/v) and

100 m l

H_{2} S O_{4}

(9.8 %w/v) are mixed then maximum mass of

B a S O_{4}

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Concentration Terms - An Introduction (w/w etc)

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