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Eshidakun
@Eshidakun
November 2021
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ДАЮ 100 БАЛЛОВ. СРОЧНО! ИТОГОВАЯ ПО ХИМИИ 8 КЛАСС. ЗА 30 МИНУТ ПОЖАЛУЙСТА
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Кляксус
1. 4Li + O2 = 2Li2O
Li2O + H2O = 2LiOH
LiOH + HNO3 = LiNO3 + H2O
2. Cu^2+ + SO4^2- + 2Na^+ + 2OH^- = Cu(OH)2↓ + 2Na^+ + SO4^2-
Cu^2+ + 2OH^- = Cu(OH)2↓
3. 8HNO3 + 3Cu = 3Cu(NO3)2 + 2NO + 4H2O
Cu - восстановитель, N - окислитель
4. SO3 + H2O = H2SO4
n (H2SO4) = n (H2O) = n (SO3) = m : M = 40 : (32 + 16 * 3) = 0,5 моль
m1 (H2O) = n * M = 0,5 * (2 + 16) =
9 г
m2 (H2O) = m (H2O) - m1 (H2O) = 420 - 9 = 411 г
m (H2SO4) = n * M = 0,5 * (2 + 32 + 16 * 4) = 49 г
w (H2SO4) = m (H2SO4) : m2 (H2O) * 100% = 49 : 411 * 100% = 11,922%
1 votes
Thanks 0
Alexei78
4Li+O2-->2Li2O
Li2O+H2O-->2LiOH
LiOH+HNO3-->LiNO3+H2O
CuSO4+2NaOH-->Cu(OH)2+Na2SO4
Cu⁺²+SO⁻²₄+2Na⁺+2OH⁻-->Cu(OH)2+2Na⁺+SO⁻²₄
Cu⁺+2OH⁻-->Cu(OH)2
HNO3+Cu-->Cu(NO3)2+NO+H2O
Cu⁰-2e-->Cu⁺¹ восстанов -ль, окисляется
N⁺⁵+3e-->N⁺² окислитель, вост - тся
8HNO3+3Cu-->3Cu(NO3)2+2NO+4H2O
дано
V(H2O) = 420 ml
m(SO3) = 40 g
---------------------------
W(H2SO4)-?
m(H2O) = V(H2O)*p(H2O = 420*1 = 420 g
n(H2O) = 420 / 18 = 23.33 mol
n(SO3) = 40 / 80 = 0.5 mol
n(SO3) = n(H2SO4) = 0.5 mol
M(H2SO4) = 98 g/mol
m(H2SO4) = n*M= 0.5 * 98 = 49 g
m(ppa H2SO4) =420+40 = 460 g
W(H2SO4) = m(H2SO4) / m(ppa H2SO4) * 100% = 49 / 460 * 100%
W(H2SO4) = 10.65%
ответ 10.65%
0 votes
Thanks 0
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Answers & Comments
Li2O + H2O = 2LiOH
LiOH + HNO3 = LiNO3 + H2O
2. Cu^2+ + SO4^2- + 2Na^+ + 2OH^- = Cu(OH)2↓ + 2Na^+ + SO4^2-
Cu^2+ + 2OH^- = Cu(OH)2↓
3. 8HNO3 + 3Cu = 3Cu(NO3)2 + 2NO + 4H2O
Cu - восстановитель, N - окислитель
4. SO3 + H2O = H2SO4
n (H2SO4) = n (H2O) = n (SO3) = m : M = 40 : (32 + 16 * 3) = 0,5 моль
m1 (H2O) = n * M = 0,5 * (2 + 16) =
9 г
m2 (H2O) = m (H2O) - m1 (H2O) = 420 - 9 = 411 г
m (H2SO4) = n * M = 0,5 * (2 + 32 + 16 * 4) = 49 г
w (H2SO4) = m (H2SO4) : m2 (H2O) * 100% = 49 : 411 * 100% = 11,922%
Li2O+H2O-->2LiOH
LiOH+HNO3-->LiNO3+H2O
CuSO4+2NaOH-->Cu(OH)2+Na2SO4
Cu⁺²+SO⁻²₄+2Na⁺+2OH⁻-->Cu(OH)2+2Na⁺+SO⁻²₄
Cu⁺+2OH⁻-->Cu(OH)2
HNO3+Cu-->Cu(NO3)2+NO+H2O
Cu⁰-2e-->Cu⁺¹ восстанов -ль, окисляется
N⁺⁵+3e-->N⁺² окислитель, вост - тся
8HNO3+3Cu-->3Cu(NO3)2+2NO+4H2O
дано
V(H2O) = 420 ml
m(SO3) = 40 g
---------------------------
W(H2SO4)-?
m(H2O) = V(H2O)*p(H2O = 420*1 = 420 g
n(H2O) = 420 / 18 = 23.33 mol
n(SO3) = 40 / 80 = 0.5 mol
n(SO3) = n(H2SO4) = 0.5 mol
M(H2SO4) = 98 g/mol
m(H2SO4) = n*M= 0.5 * 98 = 49 g
m(ppa H2SO4) =420+40 = 460 g
W(H2SO4) = m(H2SO4) / m(ppa H2SO4) * 100% = 49 / 460 * 100%
W(H2SO4) = 10.65%
ответ 10.65%