вычислите m соли, образующейся при взаимодействии 60 (г) 10% раствора уксусной кислоты с натрий гидроксидом

даноm(ppa CH3COOH) = 60gW(CH3COOH) = 10%--------------------------------m(CH3COONa)-?m(CH3COOH) = 60*10%/100%= 6 gM(CH3COOH) = 60 g/moln(CH3COOH) = m/M = 6/60 = 0.1 moln(CH3COOH) = n(CH3COONa) = 0.1 molM(CH3COONa) = 82 g/molm(CH3COONa) = n*M = 0.1 *82 = 8.2 gответ 8.2 г

Дано:m (CH3COOH p-p) = 60гW (CH3COOH) = 10% = 0.1Найти:m (CH3COONa) - ?Решение:CH3COOH + NaOH = CH3COONa + H2Om (CH3COOH) = W (CH3COOH) * m (CH3COOH p-p) = 0.1 * 60 = 6гn (CH3COOH) = m/M = 6 / 60 = 0.1 мольn (CH3COONa) = n (CH3COOH) = 0.1 моль (по уравнению)m (CH3COONa) = n * M = 0.1 * 82 = 8.2гОтвет: m (CH3COONa) = 8.2г