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Limiting Reagent or Reactant

The minimum q...

Question

The minimum quantity of $H_{2} S$ needed to precipitate 63.5 g of $C u^{+ 2}$ will be nearly:

63.5 g

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31.7 g

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34 g

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20 g

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Solution

The correct option is D 34 g
The min. quantity of $H_{2} S$ needed to precipitate $63.5 g$ of ${C u}^{+ 2}$ will be ?
$1$ mole of ${C u}^{+ 2} ⟶ 63.5 g$
Molar Mass of Cu $⟶ 63.5 g$
The reaction is
${C u}^{+ 2} + H_{2} S ⟶ C u S + 2 H^{+}$
From the reaction
$1$ mole of ${C u}^{+ 2}$ reacts with $1$ mole of $H_{2} S$
Molar mass of ${C u}^{+ 2}$ = $63.5 g {m o l}^{- 1}$
Molar mass of $H_{2} S = 34 g$
$63.5 g o f {C u}^{+ 2} ⟶ 1 m o l e o f H_{2} S ⟶ 34 g o f H_{2} S (1 m o l e o f H_{2} S = 34 g)$
Therefore, the min quantity of $H_{2} S$ needed to precipitate $63.5 g$ of ${C u}^{+ 2}$ will be $34 g$

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