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nikitakarpovich2017
@nikitakarpovich2017
November 2021
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Помогите пожалуйста очень надо
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Alexei78
Дано
m(ppa AgNO3) = 200 g
W(AgNO3) = 17%
------------------------------
m(Ag)-?
m(AgNO3) = 200*17% / 100% = 34 g
2AgNO3+Fe-->Fe(NO3)2+2Ag
M(AgNO3) = 170 g/mol
n(AgNO3) = m/M = 34/170 = 0.2 mol
2n(AgNO3) = 2n(Ag) = 0.2 mol
M(Ag) = 108 g/mol
m(Ag) = n*M = 0.2*108 = 21.6 g
ответ 21.6 г
3)
4Fe+3O2-->2Fe2O3
Fe2O3+3H2O-->2Fe(OH)3
2Fe(OH)3-->Fe2O3+3H2O
2Fe2O3+3C-->4Fe+3CO2
0 votes
Thanks 0
ЯковПервый
4 задание:
Дано:
m (AgNO3 p-p) = 200г
W (AgNO3) = 17% = 0.17
Найти:
m (Ag) - ?
Решение:
2AgNO3 + Fe = Fe(NO3)2 + 2Ag
m (AgNO3) = W (AgNO3) * m (AgNO3 p-p) = 200 * 0.17 = 34г
n (AgNO3) = m/M = 34 / 170 = 0.2 моль
n (Ag) = n (AgNO3) = 0.2 моль (по уравнению)
m (Ag) = n * M = 0.2 * 108 = 21.6г
Ответ: m (Ag) = 21.6г
5 задание:
4Fe + 3O2 = 2Fe2O3
Fe2O3 + 6HCl = 2FeCl3 + 3H2O
FeCl3 + 3KOH = Fe(OH)3 + 3KCl
2Fe(OH)3 =(t°)= Fe2O3 + 3H2O
Fe2O3 + 3CO = 2Fe + 3CO2
2 votes
Thanks 1
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Answers & Comments
m(ppa AgNO3) = 200 g
W(AgNO3) = 17%
------------------------------
m(Ag)-?
m(AgNO3) = 200*17% / 100% = 34 g
2AgNO3+Fe-->Fe(NO3)2+2Ag
M(AgNO3) = 170 g/mol
n(AgNO3) = m/M = 34/170 = 0.2 mol
2n(AgNO3) = 2n(Ag) = 0.2 mol
M(Ag) = 108 g/mol
m(Ag) = n*M = 0.2*108 = 21.6 g
ответ 21.6 г
3)
4Fe+3O2-->2Fe2O3
Fe2O3+3H2O-->2Fe(OH)3
2Fe(OH)3-->Fe2O3+3H2O
2Fe2O3+3C-->4Fe+3CO2
Дано:
m (AgNO3 p-p) = 200г
W (AgNO3) = 17% = 0.17
Найти:
m (Ag) - ?
Решение:
2AgNO3 + Fe = Fe(NO3)2 + 2Ag
m (AgNO3) = W (AgNO3) * m (AgNO3 p-p) = 200 * 0.17 = 34г
n (AgNO3) = m/M = 34 / 170 = 0.2 моль
n (Ag) = n (AgNO3) = 0.2 моль (по уравнению)
m (Ag) = n * M = 0.2 * 108 = 21.6г
Ответ: m (Ag) = 21.6г
5 задание:
4Fe + 3O2 = 2Fe2O3
Fe2O3 + 6HCl = 2FeCl3 + 3H2O
FeCl3 + 3KOH = Fe(OH)3 + 3KCl
2Fe(OH)3 =(t°)= Fe2O3 + 3H2O
Fe2O3 + 3CO = 2Fe + 3CO2