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Stoichiometric Calculations

The mass of ...

Question

The mass of $70 %$ $H_{2} S O_{4}$ required for neutralization of $1$ mole of $N a O H$ is :

70 g

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35 g

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30 g

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95 g

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Solution

The correct option is D $70 g$
$N a O H + H_{2} S O_{4} ⟶ N a_{2} S O_{4} + 2 H_{2} O$
$2 m o l 1 m o l$
$1 m o l 0.5 m o l = 49 g H_{2} S O_{4} (p u r e)$
Thus, $70 %$ $H_{2} S O_{4} r e q u i r e d = 49 \times \frac{100}{70} = 70 g$ .

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Mass of $70$ % $H_{2} S O_{4}$ solution (by mass) required for neutralisation of $1$ mol of $N a O H$ is:

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