Предмет: Химия, автор: IIslam77

Сколько грамм осадка выпадет при взаимодействии 17 г нитрата серебра с 40 г соляной кислоты?

Ответы

Автор ответа: perhydrol

0

Ответ:

AgNO3 + HCl = AgCl + HNO3

М (AgNO3) = 170г/моль

М (НСl) = 36,5г/моль

М (АgCl) = 143,5г/моль

n (AgNO3) = 17/170=0,1 моль

n (НСl) = 40/36.5=1,1 моль

Соляная кислота в избытке, расчет проводим по нитрату серебра.

n (АgCl) = n (AgNO3)

m (АgCl) = 0.1 × 143.5 = 14,35г

ответ: 14,35 г

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