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1.00 xx 10^(-3) moles of Ag^(+) and 1.0...

`1.00 xx 10^(-3)` moles of `Ag^(+)` and `1.00 xx 10^(-3)` moles of `CrO_(4)^(2-)` react together to form solid `Ag_2 CrO_4`. Calculate the amount of `Ag_2 CrO_4` formed. `(Ag_2 CrO_4 = 331.73 g mol^(-1))`

A

`0.268 g`

B

`0.166 g`

C

`0.212 g`

D

`1.66 g`

Text Solution

Verified by Experts

The correct Answer is:
B
The reaction is `2Ag^(+) + CrO_(4)^(2-) to Ag_2 CrO_4` Using the limiting reagent concept, amount of `Ag_2 CrO_4`
`= 0.5 xx 10^(-3) xx 33.1 73 = 0.166 g `
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