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Concentration Terms - An Introduction (w/w etc)

Calculate the...

Question

Calculate the amount of precipitate of $B a S O_{4}$ formed when 30 mL of 0.1 M $B a C l_{2}$ is mixed with 40 mL of 0.2 M $A l_{2} (S O_{4})_{3}$ .
(Molar mass of barium = 137 g/mol, Molar mass of Al = 27 g/mol).

0.7 g

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0.5 g

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1.2 g

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1.8 g

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Solution

The correct option is A 0.7 g
Balanced chemical reaction:
$m o l e s b e f o r e R e a c t i o n 3 B a C l_{2} 3 \times 10^{- 3} + A l_{2} (S O_{4})_{3} 8 \times 10^{- 3} \to 3 B a S O_{4} 0 + 2 A l C l_{3} 0$
$m o l e s a f t e r R e a c t i o n 3 B a C l_{2} 0 + A l_{2} (S O_{4})_{3} 7 \times 10^{- 3} \to 3 B a S O_{4} 3 \times 10^{- 3} + 2 A l C l_{3} 2 \times 10^{- 3}$

$N u m b e r o f m o l e s o f B a S O_{4} f o r m e d = 3 \times 10^{- 3} m o l$
$M a s s o f p r e c i p i t a t e f o r m e d = m o l e \times m o l a r m a s s = 3 \times 10^{- 3} \times 233 = 0.699 g \approx 0.7 g$

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