К 90г 20% раствора CH3COOH добавили 20г KHCO3. Раствор выпарили. К сухому остатку...

$1.\ n(CH_{3}COOH) = \frac{m}{Mr} = \frac{m*w}{Mr} = \frac{18}{60} = 0.3\ mol\\ \\ 2.\ n(KHCO_{3}) = \frac{20}{100} = 0.2\ mol\\ \\ ==0.3\ mol====0.2\ mol=======[0.2\ mol]====\\ 3.\ CH_{3}COOH + KHCO_{3} = CO_{2} + CH_{3}COOK + H_{2}O\\ \\ ==0.2\ mol========[[0.2\ mol]]===\\ 4. CH_{3}COOK + KOH = CH_{4} + K_{2}CO_{3}\\ \\ 5. V_{CH_{4}} = n*Vm = 0.2 *22.4 =4.48\ dm^3$

Ответ: 4.48 дм^3