h2s + HNO3 -->s+ no+H2O balance by oxidation number change
Eryy
Man
To balance the given chemical equation (h2s + HNO3 -> s + no + H2O) using the oxidation number change method, we need to assign the oxidation numbers to each element in the equation, calculate the changes in oxidation numbers, and then balance the equation accordingly.
Here's how you can do it step-by-step:
Step 1: Assign oxidation numbers
Assign oxidation numbers to each element in the equation based on a set of predetermined rules:
- Hydrogen (H) has an oxidation number of +1 when combined with nonmetals.
- Oxygen (O) has an oxidation number of -2, except in peroxides where it is -1.
- Nitrogen (N) has an oxidation number of -3 when combined with three hydrogen atoms (HNO3).
- Sulfur (S) has an oxidation number of -2 (H2S).
- The oxidation numbers should sum up to zero for a neutral compound or the charge for an ion.
Using these rules, we can assign the following oxidation numbers:
H2S: H = +1, S = -2
HNO3: H = +1, N = +5, O = -2
Step 2: Determine the change in oxidation numbers
In this reaction, sulfur (S) is going from -2 to 0, and nitrogen (N) is going from +5 to +2. The change in oxidation numbers for sulfur is +2, and for nitrogen, it is -3.
Step 3: Balance the equation
To balance the equation, we need to ensure that the total increase in oxidation numbers is equal to the total decrease in oxidation numbers. In this case, the change in oxidation numbers is not balancing, so we need to balance it by multiplying some species by appropriate coefficients.
To balance the sulfur (S) atoms, we need to multiply H2S on the left side by 4 as it has a coefficient of 1, and this will give us 4 H2S molecules that contain 4 sulfur atoms:
4 H2S + HNO3 -> s + no + H2O
Now, let's balance the nitrogen (N) atoms. To do this, we need to multiply HNO3 by 2 on the right side, so that we have 2 nitrogen atoms:
4 H2S + 2 HNO3 -> s + no + H2O
Finally, let's balance the hydrogen (H) and oxygen (O) atoms. We can see that there are 8 hydrogen atoms on the left side, but only 2 on the right side. To balance it, we need to multiply H2O on the right side by 4, giving us 8 hydrogen atoms:
4 H2S + 2 HNO3 -> s + no + 4 H2O
Now the hydrogen atoms are balanced. We also need to balance the oxygen atoms. We have 12 oxygen atoms on the right side (from 2 HNO3 and 4 H2O), but only 2 on the left side. To balance it, we need to multiply H2S by 6, giving us 12 oxygen atoms:
6 H2S + 2 HNO3 -> s + no + 4 H2O
The final balanced equation by oxidation number change method is:
6 H2S + 2 HNO3 -> S + NO + 4 H2O
S changes from -2 in H2S to zero in S.
N changes from +5 in HNO3 to +2 in NO.