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
A
`0.70`
B
`0.50`
C
`0.20`
D
`0.10`
Text Solution
Verified by Experts
The correct Answer is:
D
The balanced chemical reaction is `3BaCl_(2) + 2Na_(3)PO_(4) rarr Ba_(3)(PO_(4))_(2) + 6NaCl` IN this reaction, 3 moles of `BaCl_(2)` combines with 2 moles `Na_(3)PO_(4)`. Hence, `0.5` mole of `BaCl_(2)` require `(2)/(3) xx 0.5 = 0.33` mole of `Na_(3)PO_(4)`. Since, available `Na_(3)PO_(4) (0.2 "mole")` is less than required mole `(0.33)`, it is the limiting reactant and would determine the amount of products `Ba_(3)(PO_(4))_(2)`. `because` 2 moles of `Na_(3)PO_(4)` gives 1 mole `Ba_(3)(PO_(4))_(2)` `:. 0.2` moles of `Na_(3)PO_(4)` would give `(1)/(2) xx 0.2` `= 0.1` mole `Ba_(3)(PO_(4))_(2)`
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