Consider the reaction: `P_(4)(s)+F_(2)(g)rarrPF_(3)(g)` How many gram of `F_(2)` are needed to produce 11.2 L of `PF_(3)` at NTP?
Consider the reaction:
`P_(4)(s)+F_(2)(g)rarrPF_(3)(g)`
How many gram of `F_(2)` are needed to produce 11.2 L of `PF_(3)` at NTP?
A. 28.5gm
B. 48gm
C. 57gm
D. 85.5gm
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`P_(4)+6F_(2)rarr4PF_(3)`
`11.2` lit of `PF_(3)rArr0.5` mole of `PF_(3)`
mole of `F_(2)` required `=(6)/(4)xx0.5=(3)/(4)` moles
mass of `F_(2)` required `=(3)/(4)xx38=28-5gm`
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