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PF(3) reacts with XeF(4) to give PF(5) ...

`PF_(3)` reacts with `XeF_(4)` to give `PF_(5)`
`underset((g))(2PF_(3))+underset((s))(XeF_(4))rarrunderset((g))(2PF_(5))+underset((g))(Xe)`
If 100.0gm of `PF_(3)` and 50.0 gm `XeF_(4)` react, then which of the following statement is true?

A

`XeF_(4)` is the limiting reagent

B

`PF_(3)` is the limiting reagent

C

1.127 mol of `PF_(5)` are produced

D

0.382 mol of `PF_(5)` are produced

Text Solution

Verified by Experts

The correct Answer is:
A, D
`2PF_(3)+XeF_(4)rarr2PF_(5)+Xe`
No. of moles of `PF_(3)=(100)/(88)=1.13`
No. of moles of `XeF_(4)=(50)/(207)=0.24`
1 mol `XeF_(4)` produce 2 mol `PF_(5)`
`therefore 0.24` mol `XeF_(4)` produce `0.24xx2`
= 0.48 mol PF
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