Write a balanced equation for the reaction between rhombic and hot nitric acid
rhombic what? Is that rhomic sulfur? There is no element nor compound with rhombic as a name.
sorry its rhombic sulphur
With the hot nitric acid, S8 will oxidize from zero to +6
S + 6 HNO3 >>>H2SO4 +6NO2+2H2O When temperature of nitric acid is increased, sulfur can be oxidized to its +6 oxidation state.
To write a balanced equation for the reaction between rhombic (or crystalline) sulfur and hot nitric acid, we need to understand the chemical formula for each compound involved.
Rhombic sulfur is composed of S8 molecules, and its chemical formula is S.
Nitric acid is composed of one nitrogen atom (N) and three oxygen atoms (O), and its chemical formula is HNO3.
Now, let's proceed with writing the balanced equation using the following steps:
Step 1: Identify the reactants and products.
The reactants in this reaction are rhombic sulfur (S) and hot nitric acid (HNO3).
The products are sulfur dioxide (SO2), water (H2O), and nitrogen dioxide (NO2).
Step 2: Write the initial unbalanced equation.
S + HNO3 → SO2 + H2O + NO2
Step 3: Balance the equation by adding coefficients.
Since we have one sulfur atom on the left side and one on the right side, sulfur is already balanced.
Next, we balance the nitrogen atoms: We currently have one nitrogen atom on the right side, so we need to balance it by placing a coefficient of 2 in front of HNO3.
S + 2HNO3 → SO2 + H2O + NO2
Now, let's balance the hydrogen and oxygen atoms:
- We have six hydrogen atoms on the right side, so we balance them by placing a coefficient of 3 in front of H2O.
- We have five oxygen atoms on the left side and seven on the right side, so we balance them by placing a coefficient of 7/2 in front of NO2.
S + 2HNO3 → SO2 + 3H2O + (7/2)NO2
To remove the fractional coefficient, we can multiply all coefficients by 2, resulting in the final balanced equation:
2S + 4HNO3 → 2SO2 + 6H2O + 7NO2
This is the balanced equation for the reaction between rhombic sulfur and hot nitric acid.