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Limiting Reagent or Reactant

0.5 moles of ...

Question

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

0.2 mol

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0.5 mol

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0.4 mol

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1.5 mol

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Solution

The correct option is A 0.2 mol
The given balanced equation is, $H_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + 2 H_{2} O$
1 mol of $H_{2} S O_{4}$ reacts with 1 mol of $C a (O H)_{2}$ to give 1 mol of $C a S O_{4} .$
The amount of $H_{2} S O_{4} = 0.5$ mol
The amount of $C a (O H)_{2} = 0.2$ mol
Thus $C a (O H)_{2}$ is the limiting reagent.
Hence, the maximum number of moles of $C a S O_{4}$ formed in the reaction is 0.2 mol.

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