А) CuCl2 + Zn = ZnCl2 + Cu
б) Zn + Cl2 = ZnCl2
в) Na2O + CO2 = Na2CO3
2HNO₃ + Cu(OH)₂ = Cu(NO₃)₂ + 2H₂O1)Cu²⁺ + 2OH⁻ + 2H⁺ + 2NO₃⁻ = Cu²⁺ +2NO₃⁻ + 2H₂0
2)OH⁻ + H⁺ = H₂<span>0
</span>
<span>AgNO</span>₃ + NaCl = NaNO₃ + AgCl↓ 1)Ag⁺ + NO₃⁻ + Na⁺ + Cl⁻ = Na⁺ + NO₃⁻ + AgCl↓ <span>2)Ag</span><span>⁺</span><span> </span><span>+</span><span> </span><span>Cl</span><span>⁻</span><span> </span><span>=</span><span> </span><span>Ag</span><span>Cl</span><span>↓</span><span>
</span>
Na₂SO₃ +
2H₂SO₄ = 2NaHSO₄ + SO₂↑ + H₂O
1)2Na⁺ +
SO₃²⁻ + 4H⁺ + 2SO₄²⁻ = 2Na⁺ + 2HSO₄⁻ + SO2↑ + H2O
<span>2)SO</span><span>₃</span><span>²</span><span>⁻</span><span> + 4H</span><span>⁺</span><span> + 2SO</span><span>₄</span><span>²</span><span>⁻</span><span> = 2HSO</span><span>₄</span><span>⁻</span><span> + SO</span>₂<span>↑ + H</span>₂<span>O</span>
CuSO4+2KOH=K2SO4+Cu(Oh)2
n(CuSO4)=16г/160г/моль=0.1моль (недостаток)
n(KOH)=12г/56г/моль=0,2моль (избыток)
Релаем задачу по недостатку:
n(CuSO4)=n(Cu(OH)2)=0.1моль
m(Cu(OH)2)=0,1моль*98г/моль=9,8г
Ответ: 9,8г ; KOH в избытке
VCH4=1000·0.75=750L
VC2H6=1000·0.15=150L
VC2H4=1000·0.1=100L
750L xL
1.CH4+ 2O2 =CO2+2H2O
22.4L 44.8L
x=750·44.8/22.4=1500L
150L xL
2. C2H6+ 3.5O2 = 2CO2+3H2O
22.4L 3.5·2.4L
x=150·78.4/22.4=525L
100L xL
3.C2H4+ 3O2 = 2CO2+2H2O
22.4L 3·22.4L
x=100·67.2/22.4=300L
V(O2)=1500+525+300=2325L