1.

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

Answer»

`0.70`
`0.50`
`0.20`
`0.10`

Solution :`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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