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, isA. `0.70`B. `0.50`C. `0.20`D. `0.10`

If 0.50 mol of `CaCl_(2)` is mixed with 0.20 mol of `Na_(3)PO_(4)`, th

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, isA. `0.70`B. `0.50`C. `0.20`D. `0.10`
Answer» Correct Answer - 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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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, isA. `0.70`B. `0.50`C. `0.20`D. `0.10`

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