1. «Четвертый лишний» Из данной группы выберите формулу лишнего вещества А) SiO2, CO2, SO3 , SnO B) Na2O, Li2O, CaO, CuO C) MgO, CuO, K2O, Mn2O7 А) В) С) [3] 2. Установите соответствие: Исходные вещества Продукты реакции 1. Al(OH)3+3HCl→ А. Al2O3+3H2O 2. 2Al(OH)3→ B.AlCl3+3H2O 3. AlCl3+3KOH→ C.3KCl+Al(OH)3

Объяснение:

1. «Четвертый лишний» Из данной группы выберите формулу лишнего вещества

А) SnO

B) CuO

C) Mn2O7

2. Установите соответствие:

1В 2А 3С

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1) Находим молярные массы и количество молей в-в, вступающих в реакцию и хлорида магния:

М(MgO) = 24 + 16 = 40 г/моль

n(MgO) = m/M = 120/40 = 3 моль

М(HCl) = 1 + 35,5= 36,5 г/моль

n(HCl) = m/M = 210/36,5 = 5,7 моль

М(MgC₂l) = 24 + 2*35,5 =  95 г/моль

2) Уравнение реакции:

MgO + 2HCl = MgCl₂ + H₂O

По уравнению реакции видно что магний оксид и хлороводород вступают в реакцию в соотношении:  MgO : HCl = 1 : 2

Значит в данной реакции HCl  в недостатке и по ней ведем расчеты.

3) По уравнению реакции составляем пропорцию:

2*36,5 г HCl 95 г MgCl₂

   210 г HCl  X г MgCl₂

X = 210*95/(2*36,5) = 273,3 = 273 г

ответ: 273 г