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Laws of Mass Conservation

Calculate the...

Question

Calculate the mass of hydrated copper sulphate required to produce 56.90 g of the complex compound assuming other reagents to be present in excess.
$C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O$
$C u S O_{4} + 2 K C N \to C u (C N)_{2} + K_{2} S O_{4}$
$2 C u (C N)_{2} \to 2 C u C N + (C N)_{2}$
$2 C u C N + 6 K C N \to 2 K_{3} [C u (C N)_{4}]$

$(Molar mass of K = 39 g / m o l, C u = 63.5 g / m o l)$

122.4 g

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19.96 g

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12.24 g

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49.9 g

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Solution

The correct option is D 49.9 g
Let the weight of $C u S O_{4} .5 H_{2} O$ be $M g$ .
Since, atoms of $C u$ are conserved, apply POAC on $C u$ ,
$1 \times moles of C u S O_{4} .5 H_{2} O = moles of K_{3} [C u (C N)_{4}]$

$\Rightarrow \frac{weight of C u S O_{4} .5 H_{2} O}{molar mass} = \frac{weight of complex}{molar mass of complex}$

$\Rightarrow \frac{M}{249.5} = \frac{56.90}{284.50}$

On solving, $M = 49.9 g$

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Similar questions

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O

C u S O_{4} + 2 K C N \to C u (C N)_{2} + K_{2} S O_{4}

2 C u (C N)_{2} \to 2 C u C N + (C N)_{2}

2 C u C N + 6 K C N \to 2 K_{3} [C u (C N)_{4}]

(Molar mass of K = 39 g / m o l, C u = 63.5 g / m o l)

C u S O_{4} .5 H_{2} O

K C N

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O (Yield = 100%)

C u S O_{4} + 2 K C N \to C u (C N)_{2} + K_{2} S O_{4} (Yield = 80%)

2 C u (C N)_{2} \to 2 C u C N + (C N)_{2} (Yield = 60%)

2 C u C N + 6 K C N \to 2 K_{3} [C u (C N)_{4}] (Yield = 50%)

C u S O_{4} .5 H_{2} O

2

(C u S O_{4} .5 H_{2} O)

C u = 63.5 g/mol

C u S O_{4} .5 H_{2} O

K C N

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O (Yield = 100%)

C u S O_{4} + 2 K C N \to C u (C N)_{2} + K_{2} S O_{4} (Yield = 80%)

2 C u (C N)_{2} \to 2 C u C N + (C N)_{2} (Yield = 60%)

2 C u C N + 6 K C N \to 2 K_{3} [C u (C N)_{4}] (Yield = 50%)

C u S O_{4} .5 H_{2} O

[C u S O_{4} .5 H_{2} O]

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O

C u = 64, S = 32, O = 16, H = 1

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