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Potassium super oxide, KO_(2) is an air purifier as it eacts with CO_(2) and releases O_(2),4KO_(2)(s)+2CO_(2)(g) to 2K_(2)CO_(3)(s)+3O_(2)(g) Calculate the mass of KO_(2) neeeded to react with 50 L of CO_(2) at 25^(@)C, 1 atm. |
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Answer» Solution :STRATEGY: * Find `V_(m)` at given (T,P) * USE `n=(V)/(V_(m))` to find n of `CO_(2)`. * Use `m_(KO_(2))=n_(KO_(2)):` * `n_(KO_(2))` is obtained from stoichiometry. * `V_(m)(CO_(2))=(RT)/(P)=0.0821xx298.15//1L=24.47" L "mol^(-1)`: * `n_(CO_(2))=(V)/(V_(m))=(50L)/(24.47" L "mol^(-1))=2.043` mole * From stoichiometry, 2 mol `CO_(2)` reacts with 4 mol `KO_(2)therefore n_(KO_(2))=(4)/(2)xxn_(CO_(2))=2xx2.043` mol =4.086 mol * FINALLY use m=nM or, `m_(KO_(2))=4.086molxx71.1g" "mol^(-1)=290.56g` |
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