Question

Помогите!!!!! Расставить коэффициенты в уровнениях методом полуреакций Na2SO4+C=Na2S+CO+H2O Na2Cr2O7+NaNO2+H2SO4=NaNO3+Cr2(SO4)+Na2SO4+H2O S+KOH=K2SO3+K2S+H2O

Answer 1

Na2SO4+C=Na2S+CO+H2O( ну вот как у тебя справа получилась вода, если слева нет водорода?) 

Na2Cr2O7+5NaNO2+2H2SO4=5NaNO3+Cr2(SO4)+Na2SO4+2H2O 

S+KOH=K2SO3+K2S+H2O( то же самое)

Answer 2

Balancing Equations using Half-Reaction Method

To balance the given equations using the half-reaction method, we need to follow these steps:

1. Split the equation into two half-reactions: oxidation and reduction. 2. Balance the atoms in each half-reaction, excluding oxygen and hydrogen. 3. Balance the oxygen atoms by adding water molecules (H2O) to the side that needs more oxygen. 4. Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen. 5. Balance the charge by adding electrons (e-) to the side that needs more negative charge. 6. Multiply the half-reactions by appropriate coefficients to equalize the number of electrons transferred. 7. Add the balanced half-reactions together and cancel out any common terms.

Let's balance the given equations step by step:

Equation 1: Na2SO4 + C = Na2S + CO + H2O

First, let's split the equation into two half-reactions:

Oxidation half-reaction: C → CO Reduction half-reaction: Na2SO4 → Na2S

Now, let's balance the atoms in each half-reaction:

Oxidation half-reaction: C → CO Reduction half-reaction: Na2SO4 → Na2S + O

Next, let's balance the oxygen atoms by adding water molecules:

Oxidation half-reaction: C → CO + H2O Reduction half-reaction: Na2SO4 → Na2S + O2

Now, let's balance the hydrogen atoms by adding hydrogen ions:

Oxidation half-reaction: C + 2H2O → CO + 4H+ + 4e- Reduction half-reaction: Na2SO4 → Na2S + O2

Next, let's balance the charge by adding electrons:

Oxidation half-reaction: C + 2H2O → CO + 4H+ + 4e- Reduction half-reaction: Na2SO4 + 4e- → Na2S + O2

To equalize the number of electrons transferred, we need to multiply the oxidation half-reaction by 4:

Oxidation half-reaction: 4C + 8H2O → 4CO + 16H+ + 16e- Reduction half-reaction: Na2SO4 + 4e- → Na2S + O2

Now, let's add the balanced half-reactions together:

4C + 8H2O + Na2SO4 + 4e- → 4CO + 16H+ + 16e- + Na2S + O2

Canceling out the common terms, we get the balanced equation:

Final balanced equation: 4C + 8H2O + Na2SO4 → 4CO + 16H+ + Na2S + O2

Equation 2: Na2Cr2O7 + NaNO2 + H2SO4 = NaNO3 + Cr2(SO4) + Na2SO4 + H2O

Let's balance this equation using the same steps as before:

Oxidation half-reaction: Na2Cr2O7 → Cr2(SO4) Reduction half-reaction: NaNO2 → NaNO3

Balancing the atoms in each half-reaction:

Oxidation half-reaction: Na2Cr2O7 → Cr2(SO4) + O Reduction half-reaction: NaNO2 → NaNO3

Balancing the oxygen atoms by adding water molecules:

Oxidation half-reaction: Na2Cr2O7 → Cr2(SO4) + 7H2O Reduction half-reaction: NaNO2 → NaNO3

Balancing the hydrogen atoms by adding hydrogen ions:

Oxidation half-reaction: Na2Cr2O7 + 14H+ → Cr2(SO4) + 7H2O Reduction half-reaction: NaNO2 → NaNO3

Balancing the charge by adding electrons:

Oxidation half-reaction: Na2Cr2O7 + 14H+ + 6e- → Cr2(SO4) + 7H2O Reduction half-reaction: NaNO2 + 2e- → NaNO3

To equalize the number of electrons transferred, we need to multiply the reduction half-reaction by 6:

Oxidation half-reaction: Na2Cr2O7 + 14H+ + 6e- → Cr2(SO4) + 7H2O Reduction half-reaction: 6NaNO2 + 12e- → 6NaNO3

Adding the balanced half-reactions together:

Na2Cr2O7 + 14H+ + 6e- + 6NaNO2 + 12e- → Cr2(SO4) + 7H2O + 6NaNO3

Canceling out the common terms, we get the balanced equation:

Final balanced equation: Na2Cr2O7 + 6NaNO2 + 14H+ + H2SO4 → Cr2(SO4) + 7H2O + 6NaNO3 + Na2SO4

Equation 3: S + KOH = K2SO3 + K2S + H2O

Let's balance this equation using the same steps as before:

Oxidation half-reaction: S → K2SO3 Reduction half-reaction: KOH → K2S

Balancing the atoms in each half-reaction:

Oxidation half-reaction: S → K2SO