Предмет: Химия, автор: alekcander555

Срочно 99 баллов .найти массу (cus ) в г при взаимодействии 15 г Na2S и 20г Cu(NO3)2

Ответы

Автор ответа: Ботаник122
0
Для начала найдём кол-во вещества обоих реагентов:
v(
Na2S)=m/M=15/78≈0.19 моль.
v(Cu(NO3)2)=m/M=20/188≈0.106 моль.

так как кол-во вещества Na2S меньше, то и будем решать через него.
                        15                              х г
Cu(NO3)2 + Na2S = 2NaNO3 + CuS↓
                     78г/моль                  96г/моль
 
Найдём Х через пропорцию:
х г=15*96/78≈18.46 грамм.

Ответ: 18.46 грамм.
Автор ответа: Ботаник122
0
что-то я ошибся в расчётах)
Автор ответа: ablyakimovtima
0
бывает.
Автор ответа: ablyakimovtima
0
Na2S+Cu(NO3)2 = CuS + 2NaNo3
молярная масса Na2S(23*2+32=78г./моль) количество вещества Na2S(15/78=0.19моль)
молярная масса Cu(NO3)2(64+(14+48)*2)=188) количества вещества Cu(NO3)2(20/188=0.10)
кол-во вещества CuS = 0.19моль молярная масса CuS (96г/моль) Масса CuS = кол-во вещества*молярную массу=0.19*96=18.24г.
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Предмет: История, автор: Аноним