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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? |
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Answer» `XeF_(4)` is the limiting reagent 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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