You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.00025

mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.025

mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0025

mol

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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 $B a S O_{4}$

Suggest Corrections

Similar questions

10 ml of 1M Bacl2 solution and 5ml 0.5 M K2SO4 are mixed together to precipitate out BaS04 .The amount of BaSO4 precipited will be

a) 0.005.mol b) 0.00025mol

c)0.025mol. c)0.0025 mol

10

1 M B a C l_{2}

5

0.5 M K_{2} S O_{4}

B a S O_{4}

B a S O_{4}

B a {C l}_{2}

K_{2} S O_{4}

B a S O_{4}

B a S O_{4}

B a C l_{2}

A l_{2} (S O_{4})_{3}

B a S O_{4}

0.5

B a C l_{2}

1

H_{2} S O_{4}

Join BYJU'S Learning Program