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Stoichiometric Calculations

A metal M f...

Question

A metal $M$ forms the sulphate $M_{2} (S O_{4})_{3}$ . A $0.596 g$ sample of the sulphate reacts with excess $B a C l_{2}$ to give $1.220 g B a S O_{4}$ . What is the atomic weight of $M$ ? (Atomic weights: $S = 32, B a = 137.3$ )

26.9

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69.7

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55.8

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23

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Solution

The correct option is B $26.9$
Moles of $B a S O_{4} = \frac{1.22}{233.3}$
Moles of $M_{2} (S O_{4})_{3} = \frac{1.22}{233.3} \times \frac{1}{3}$ $= 1.743 \times 10^{- 3}$
Weight of $M_{2} (S O_{4})_{3} = 0.596$
$∴ 1.743 \times 10^{- 3} (2 M + 96 \times 3) = 0.596$
Hence, $M = 26.9$

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B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l

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[N a = 23, O = 16, S = 32, B a = 132]

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