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What should be the weight and moles of A...

What should be the weight and moles of AgCl precipitate obtained on adding `500 ml` of `0.20 M HCl` in `30 g` of `AgNO_(3)` solution? `(agNO_(3)=170)`

A

(a)` 14.35 g`

B

(b)` 15 g`

C

(c )`18 g `

D

(d)`19 g `

Text Solution

Verified by Experts

The correct Answer is:
A
`AgNO_(3)+HCl rarr AgCl+HNO_(3)`
`30/170 (500xx0.2)/1000`
`t=0 0.176 "mole "0.1"mole limiting" =14.345 g`
`t=t 0.076 "mole"0 0.1 "mole"`
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