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Limiting Reagent or Reactant

10 mL of 1 ...

Question

$10$ mL of $1 M$ $B a C l_{2}$ solution and $5$ mL of $0.5 M$ $K_{2} S O_{4}$ are mixed together to precipitate out $B a S O_{4}$ . The amount of $B a S O_{4}$ precipated will be :

0.005

mol

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0.00025

mol

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0.025

mol

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0.0025

mol

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Solution

The correct option is D $0.0025$ mol
$B a C l_{2} + K_{2} S O_{4} \to B a S O_{4} ↓ + 2 K C l$
$10$ mL of $1 M$ $B a C l_{2}$ and $5$ mL of $0.5 M$ $K_{2} S O_{4}$ are mixed together.
Here, $K_{2} S O_{4}$ is the limiting reagent.
$1$ mmol $K_{2} S O_{4} \equiv 1$ mmol $B a S O_{4}$
$2.5$ mmol $K_{2} S O_{4} \equiv 2.5$ mmol $B a S O_{4}$ $\equiv \frac{2.5}{1000} = 0.0025$ mol $B a S O_{4}$

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