If 0.50 mol of barium chloride (BaCl(2)...

If `0.50 mol` of barium chloride `(BaCl_(2))` is mixed with `0.20` mol of sodium phosphate `(Na_(3) PO_(4))` the maximum number of moles of barium phosphate `[Ba_(3)(PO_(4))_(2)]` formed is

`0.25`

`0.10`

`0.40`

`0.50`

Text Solution

Verified by Experts

The correct Answer is:

`underset(3 mol)(3BaCl_(2)) + underset(2 mol) (2Na_(3) PO_(4)) rarr underset(1 mol) (Ba_(3)(PO_(4))_(2)) + underset(6 mol) (6NaCl)`
`0.50 mol BaCl_(2) xx (1 mol Ba_(3) (PO_(4))_(2))/(3 mol BaCl_(2)) = 0.17 mol Ba_(3) (PO_(4))_(2)`
`0.20 mol Na_(3) PO_(4) xx (1 mol Ba_(3) (PO_(4))_(2))/(2.0 mol Na_(3) PO_(4)) = 0.1 mol Ba_(3) (PO_(4))_(2)`
Therefore, `Na_(3) PO_(4)` is the limiting rectant and the maximum moles of `Ba_(3) (PO_(4))_(2)` formed is `0.1` mol.

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If 0.50 mol of BaCl_(2) is mixed with 0.20 mol of Na_(3)PO_(4) , the maximum number of moles of Ba_(3)(PO_(4))_(2) that can be formed is

0.7

0.5

0.2

0.1

If 0.5 mol of BaCl_(2) is mixed with 0.2 mol of Na_(3)PO_(4) , the maximum number of moles of Ba_(3)(PO_(4))_(2) that can be formed is

`0.7`

`0.5`

`0.1`

`0.2`

If 0.5 mol of BaCl_(2) is mixed with 0.2 mole of Na_(3)PO_(4) , the maximum number of moles of Ba_(3)(PO_(4))_(2) that can be formed is

0.7

0.5

0.3

0.1