Question

Решите окислительно- восстановительные уравнения 1)C+H2SO4——CO2+SO2+H2O2)C+HNO3——CO2+NO+H2O3)C+H2SO4——CO2+SO2+H2O

Answer 1

2H2SO4(конц., гор.) + C(графит) = 2SO2 + CO2 + 2H2O

3C + 4 HNO3 = 3CO2 + 4 NO+ 2 H2O

Answer 2

Вот правильные ответы так должно и быть

Answer 3

1. C + H2SO4 → CO2 + SO2 + H2O

This is a redox reaction, where carbon is oxidized and sulfur is reduced. To balance the equation, we need to break it down into half-reactions and balance the electrons.

Half-reaction for oxidation: C → CO2 + 4 e-

Half-reaction for reduction: H2SO4 + 6 e- → SO2 + 2 H2O

To balance the electrons, we need to multiply the oxidation half-reaction by 6 and the reduction half-reaction by 4:

6 C + 6 H2SO4 → 6 CO2 + 6 SO2 + 6 H2O (oxidation) 4 H2SO4 + 24 e- → 4 SO2 + 8 H2O (reduction)

Finally, we can add the two half-reactions and simplify:

6 C + 10 H2SO4 → 6 CO2 + 10 SO2 + 14 H2O

2. C + HNO3 → CO2 + NO + H2O

This is also a redox reaction, where carbon is oxidized and nitrogen is reduced.

Half-reaction for oxidation: C → CO2 + 4 e-

Half-reaction for reduction: HNO3 + 3 e- → NO + 2 H2O

Again, we balance the electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 4:

3 C + 3 HNO3 → 3 CO2 + 3 NO + 6 H2O (oxidation) 4 HNO3 + 12 e- → 4 NO + 8 H2O (reduction)

Adding the two half-reactions and simplifying:

3 C + 7 HNO3 → 3 CO2 + 4 NO + 6 H2O

3. C + H2SO4 → CO2 + SO2 + H2O

This is similar to the first equation, where carbon is oxidized and sulfur is reduced.

Half-reaction for oxidation: C → CO2 + 4 e-

Half-reaction for reduction: H2SO4 + 2 e- → SO2 + 2 H2O

Multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 4 to balance the electrons:

2 C + 2 H2SO4 → 2 CO2 + 2 SO2 + 4 H2O (oxidation) 4 H2SO4 + 8 e- → 4 SO2 + 8 H2O (reduction)

Adding the two half-reactions and simplifying:

2 C + 6 H2SO4 → 2 CO2 + 6 SO2 + 12 H2O