Предмет: Химия, автор: myravei1212

2 задачи которые на скринах нужно решить
прошу помогите!!

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Ответы

Автор ответа: dobriakua
0

Задача 7.1.45

Дано:

V2(HCl) = 33,6 л

V1 р-ну(HCl) = 2 л = 2000 см3

ω1(HCl) = 10%

ρ1 р-ну(HCl) = 1,047 г/см3

Знайти:

ω3(HCl) - ?

Розв'язання:

1) m1 р-ну(HCl) = ρ1 р-ну * V1 р-ну = 1,047 * 2000 = 2094 г;

2) m1(HCl) = ω1 * m1 р-ну / 100% = 10% * 2094 / 100% = 209,4 г;

3) n2(HCl) = V2(газа) / Vm = 33,6 / 22,4 = 1,5 моль;

4) m2(HCl) = n2 * M = 1,5 * 36,5 = 54,75 г;

5) m3(HCl) = m1 + m2 = 209,4 + 54,75 = 264,15 г;

6) m3 р-ну(HCl) = m1 р-ну + m2 = 2094 + 54,75 = 2148,75 г;

7) ω3(HCl) = m3 * 100% / m3 р-ну = 264,15 * 100% / 2148,75 = 12,29%.

Відповідь: Масова частка HCl становить 12,29%.

Задача 6.2.7

Дано:

m(суміші) = 10,4 г

m(MgSO4) = 36 г

Знайти:

m(Mg) - ?

m(MgO) - ?

Розв'язання:

1) Mg + H2SO4 → MgSO4 + H2;

MgO + H2SO4 → MgSO4 + H2O;

2) Нехай m(Mg) = (х) г, тоді m(MgO) = (10,4 - х) г;

3) n(Mg) = m / M = (x / 24) моль;

4) n1(MgSO4) = n(Mg) = (x / 24) моль;

5) n(MgO) = m / M = ((10,4 - х) / 40) моль;

6) n2(MgSO4) = n(MgO) = ((10,4 - х) / 40) моль;

7) n заг.(MgSO4) = m / M = 36 / 120 = 0,3 моль;

8) n заг.(MgSO4) = n1 + n2;

0,3 = (x / 24) + ((10,4 - х) / 40);

х = 2,4;

9) m(Mg) = х = 2,4 г;

10) m(MgO) = 10,4 - х = 10,4 - 2,4 = 8 г.

Відповідь: Маса Mg становить 2,4 г; MgO - 8 г.


myravei1212: Спасибо огромное!!
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