1.

3 BaCl_(2) + 2Na_(3)PO_(4) to Ba_(3)(PO_(4))_(2)+6NaCl Maximum amount of Ba_(3)(PO_(4))_(2) formed when 2 moles each of Na_(3)PO_(4) and BaCl_(2) react is

Answer»

4 mol
1 mol
`(2)/(3) mol`
`(1)/(3) mol`

Solution :`3 BaCl_(3)+2Na_(3)PO_(4)to Ba_(3)(PO_(4))_(2)+6NaCl`
Here `BaCl_(2)` is the limiting reactant
No. of moles of `Ba_(3)(PO_(4))_(2)` formed
`=(1)/(3)xx"N0. of moles of" BaCl_(2) = (2)/(3)` mol


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