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Для сокращения записи обозначу фенол как PhOH, фенолят натрия PhONa.
Дано:
m(PhOH) = 188г
m(NaOH) = 90г
φ = 90% = 0,90
Найти:
m(PhONa)пр
Решение:
PhOH + NaOH → PhONa + H₂O
Mr(PhOH) = 94 г/моль
Mr(NaOH) = 40 г/моль
Найдем массу натрия, необходимую для реакции с 188г фенола:
m₁(NaOH) = (188×40)÷94 = 80г, а по условию 90г, значит он в избытке. Будем считать массу фенолята по фенолу:
Mr(PhONa) = 116 г/моль
m(PhONa) = (188×116)÷94 = 232г
m(PhONa)пр = 232×0,90 = 208,8г
ответ: m(PhONa)пр = 208,8г
Так что учить надо только формулы высших кислот а у - истых просто отнимаем кислород . Соляную придется выучить, или знать, что ее делают из поваренной соли NaCl