0.5 mole of $H_{2} S O_{4}$ is mixed with 0.2 mole of $C a {(O H)}_{2}$ . The maximum number of moles of $C a S O_{4}$ formed is? and Identify limiting reagent.

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0.2 mole of $C a S O_{4}$ and $C a {(O H)}_{2}$ is limiting reagent.

0.2 mole of $C a S O_{4}$ and $H_{2} S O_{4}$ is limiting reagent.

0.4 mole of $C a S O_{4}$ and $C a {(O H)}_{2}$ is limiting reagent.

1.3 mole of $C a S O_{4}$ and $H_{2} S O_{4}$ is limiting reagent.

answer is A.

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Detailed Solution

$H_{2} S O_{4} + C a {(O H)}_{2} \to C a S O_{4} + 2 H_{2} O$

One mole of $H_{2} S O_{4}$ react with one mole of $C a {(O H)}_{2}$ to form one mole of $C a S O_{4}$ . Even though there is 0.5 mole of $H_{2} S O_{4}$ only 0.2 mole of $H_{2} S O_{4}$ react. So, $H_{2} S O_{4}$ is in excess amount therefore $H_{2} S O_{4}$ is excess reagent and $C a {(O H)}_{2}$ is limiting reagent.