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If 0.50 mol of CaCl(2) is mixed with 0.2...

If 0.50 mol of `CaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, the maximum number of moles of `Ca_(3)(PO_(4))_(2)` which can be formed, is

A

`0.70`

B

`0.50`

C

`0.20`

D

`0.10`

Text Solution

Verified by Experts

The correct Answer is:
D
`3 CaCl_(2)+2 Na_(3)PO_(4)to Ca_(3)(PO_(4))_(2)+6NaCl`
`therefore` 2Moles of `Na_(3)PO_(4)=3` mole of `CaCl_(2) = 1` mole `Ca_(3)(PO_(4))_(2)`
`therefore 0.2` mole of `Na_(3)PO_(4)=0.3` mole of `CaCl_(2)=0.1` mole of `Ca_(3)(PO_(4))_(2)`.
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