$0.5$ moles of $H_{2} S O_{4}$ is mixed with $0.2$ moles of $C a {(O H)}_{2}$ . The maximum number of moles of $C a S O_{4}$ formed is :

0.2

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0.5

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0.4

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1.5

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Solution

The correct option is A $0.2$
$C a {(O H)}_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H_{2} O$
0.2 0.5 0 ---- Initial
0 0.5 - 0.2 0.2 ----- Final
Since 1 mole of $C a (O H)_{2}$ reacts with 1 mole of $H_{2} S O_{4}$ . So, $C a (O H)_{2}$ is the limiting reagent.

So, 0.2 mol of $C a (O H)_{2}$ is formed.

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$0.5$ moles of $H_{2} S O_{4}$ is mixed with $0.2$ moles of $C a {(O H)}_{2}$ . The maximum number of moles of $C a S O_{4}$ formed is :