1.

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?

Answer»

`XeF_(4)` is the limiting reagent
`PF_(3)` is the limiting reagent
1.127 mol of `PF_(5)` are produced
0.382 mol of `PF_(5)` are produced

SOLUTION :`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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