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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. 28.5gmB. 48gmC. 57gmD. 85.5gm |
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Answer» `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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