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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? |
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Answer» 28.5gm `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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