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If excess `F_(2)(g)` reacts at `0^(@)C` and 1.0 atm pressure with `Br_(2)(g)` to give a compound `BrF_(n)`, if 224ml of `Br_(2)(g)` at the same temperature and pressure produced 3.5g of `BrF_(n)`, what is n? |
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Answer» `Br_(2)+nF_(2)rarr2BrF_(n)` `224 ml Br_(2)` at 273 K and 1 atm `n_(Br_(2))=((1)(224)xx10^(-3))/(0.0821xx273)=0.01` mol `n_(Br)F_(n)=2n_(Br_(2))=0.02` mol `(3.5)/((80+19n))=0.2` 3.5=1.6+0.38n 1.9=0.38n `n=(1.9)/(0.38)=5` |
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