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A mixture containing equimolar amounts of $Ca {(OH)}_{2}$ and $Al {(OH)}_{3}$ requires 0.5 L of 4.0 M HCI to react with it completely. Moles of the mixture are:

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0.8

0.04

0.4

answer is C.

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Detailed Solution

$\begin{aligned} \underset{a}{Ca (OH)_{2}} + \underset{2 a}{2 HCl} ⟶ {CaCl}_{2} + 2 H_{2} O \end{aligned}$

$\begin{aligned} \underset{a}{Al (OH)_{3}} + \underset{3 a}{3 HCl} ⟶ {AlCl}_{3} + 3 H_{2} O \end{aligned}$

$\begin{array}{l} 5 a = 0.5 \times 4 \\ 5 a = 2.0 \\ a = \frac{2.0}{5} \end{array}$

$∴$ Total moles $= 2 a = \frac{4}{5}$

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