Предмет: Химия, автор: sl251

Сколько грамм магния может раствориться в 3,65г соляной кислоты?

Ответы

Автор ответа: Ira2207
1

Ответ:

m=? 3,65 г Mg +

2HCl = MgCl2 + H2n= 1 моль

n=2 мольМ= 24 г/моль

М= 36 г/мольm= 24 г

m = 72 гХг

Mg - 3,65 г HCl24 г Mg - 72 г HCLm(Mg) = 24 * 3,65 / 72 = 1,2 г

Автор ответа: viktorpolezaev62
0

Дано:

m(HCl)= 3,65 г

Найти:

m(Mg)-?

Решение:

2HCl+Mg=MgCl2+H2

n(HCl)=m/M= 3,65г:36,5 г/моль= 0,1 моль

M(HCl)=1+35,5 = 36,5 г/моль

n(HCl)=n(Mg)=0,05 моль

m(Mg)=n*M= 0,05 моль* 24г/моль=1,2 г

M(Mg)= 24 г/моль

Ответ: m(Mg)= 1,2 г

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