To solve the chemical equation given, we will break it down as follows:
1. Identify the Reactants and Products:
- Reactants: These are the substances that participate in the reaction. According to the given equation:
- Lead(II) sulfide (PbS)
- Hydrogen peroxide (H₂O₂)
- Products: These are the substances that are produced from the reaction. According to the given equation:
- Lead(II) sulfate (PbSO₄)
- Water (H₂O)
2. Write the chemical equation with the coefficients:
- The chemical equation given is already balanced:
[tex]\[
\text{PbS} + \text{H}_2\text{O}_2 \longrightarrow \text{PbSO}_4 + \text{H}_2\text{O}
\][/tex]
- Here, each reactant and product has a coefficient of 1, indicating the stoichiometric ratio.
3. Determine the Stoichiometric Coefficients:
- For Lead(II) sulfide (PbS): The coefficient is 1.
- For Hydrogen peroxide (H₂O₂): The coefficient is 1.
- For Lead(II) sulfate (PbSO₄): The coefficient is 1.
- For Water (H₂O): The coefficient is 1.
4. Summarize the Reactants and Products:
- Reactant stoichiometric ratios:
[tex]\[
\text{PbS} \longrightarrow 1
\][/tex]
[tex]\[
\text{H}_2\text{O}_2 \longrightarrow 1
\][/tex]
- Product stoichiometric ratios:
[tex]\[
\text{PbSO}_4 \longrightarrow 1
\][/tex]
[tex]\[
\text{H}_2\text{O} \longrightarrow 1
\][/tex]
5. Conclusion:
- The chemical reaction, balanced with stoichiometric coefficients, is:
[tex]\[
\text{PbS} + \text{H}_2\text{O}_2 \longrightarrow \text{PbSO}_4 + \text{H}_2\text{O}
\][/tex]
- This indicates that 1 mole of Lead(II) sulfide reacts with 1 mole of Hydrogen peroxide to produce 1 mole of Lead(II) sulfate and 1 mole of Water.
By examining the balanced equation, we can conclude that the stoichiometric coefficients for both the reactants and products are all 1.
