Question #4b2cc

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Question #4b2cc

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Answer 1 · 2016-12-07T05:35:14

Considering the percentage of $H C l$ $HCl$ solution as w/v the mass of $H C l$ $HCl$ present in 1 liter solution is
$= \frac{3 g}{100 m L} \times 1000 m L = 30 g$ $=(3g)/(100mL)xx1000mL=30g$

$= \frac{30 g}{36.5 \frac{g}{mol}} \approx 0.82 m o l$ $=(30g)/(36.5g/"mol")~~0.82mol$

The equation of nutralisation reaction

$H C l + N a O H \to N a C l + H_{2} O$ $HCl+NaOH->NaCl+H_2O$

This equatiion reveals that the reactants interact in $1 : 1$ $1:1$ mole ratio.

So $0.82 m o l$ $0.82 mol$ HCl requires $0.82$ $0.82$ mole NaOH for nutralisation.

So the mass of $N a O H$ $NaOH$ required for nutralisation is
$= 0.82 m o l \times 40 \frac{g}{mol} = 32.8 g$ $=0.82molxx40g/"mol"=32.8g$

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Question #4b2cc

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