вычислите m соли, образующейся при взаимодействии 60 (г) 10% раствора уксусной кислоты с натрий гидроксидом
Ответы на вопрос
Ответил Alexei78
0
дано
m(ppa CH3COOH) = 60g
W(CH3COOH) = 10%
--------------------------------
m(CH3COONa)-?
m(CH3COOH) = 60*10%/100%= 6 g
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m/M = 6/60 = 0.1 mol
n(CH3COOH) = n(CH3COONa) = 0.1 mol
M(CH3COONa) = 82 g/mol
m(CH3COONa) = n*M = 0.1 *82 = 8.2 g
ответ 8.2 г
m(ppa CH3COOH) = 60g
W(CH3COOH) = 10%
--------------------------------
m(CH3COONa)-?
m(CH3COOH) = 60*10%/100%= 6 g
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m/M = 6/60 = 0.1 mol
n(CH3COOH) = n(CH3COONa) = 0.1 mol
M(CH3COONa) = 82 g/mol
m(CH3COONa) = n*M = 0.1 *82 = 8.2 g
ответ 8.2 г
Ответил ЯковПервый
0
Дано:
m (CH3COOH p-p) = 60г
W (CH3COOH) = 10% = 0.1
Найти:
m (CH3COONa) - ?
Решение:
CH3COOH + NaOH = CH3COONa + H2O
m (CH3COOH) = W (CH3COOH) * m (CH3COOH p-p) = 0.1 * 60 = 6г
n (CH3COOH) = m/M = 6 / 60 = 0.1 моль
n (CH3COONa) = n (CH3COOH) = 0.1 моль (по уравнению)
m (CH3COONa) = n * M = 0.1 * 82 = 8.2г
Ответ: m (CH3COONa) = 8.2г
m (CH3COOH p-p) = 60г
W (CH3COOH) = 10% = 0.1
Найти:
m (CH3COONa) - ?
Решение:
CH3COOH + NaOH = CH3COONa + H2O
m (CH3COOH) = W (CH3COOH) * m (CH3COOH p-p) = 0.1 * 60 = 6г
n (CH3COOH) = m/M = 6 / 60 = 0.1 моль
n (CH3COONa) = n (CH3COOH) = 0.1 моль (по уравнению)
m (CH3COONa) = n * M = 0.1 * 82 = 8.2г
Ответ: m (CH3COONa) = 8.2г
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