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To 10 mL of 1 M BaCl(2) solution 5mL of...

To `10 mL` of `1 M BaCl_(2)` solution `5mL` of `0.5 M K_(2)SO_(4)` is added. `BaSO_(4)` is precipitated out. What will happen?

A

0.005 mol

B

0.00025 mol`

C

0.025 mol

D

0.0025 mol

Text Solution

Verified by Experts

The correct Answer is:
D
`underset({:(10 xx 1),(= 10 "mol"):})(BaCl_(2)) + underset({:(5 xx 0.5),(= 2.5 "mmo"):})(K_(2)SO_(4)) rarr (BaSO_(4) darr) + 2 KCl`
Here, `K_(2)SO_(4)`is the limiting reagent
1 mmol `K_(2) SO_(4) -= 1"mmol" BaSO_(4)`
`2.5 "mmol" K_(2) SO_(4) -= 2.5 "mmol" BaSO_(4)`
`-= (2.5)/(1000) = 0.0025 "mol" BaSO_(4)`
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