Complete the following equations
(i) `PbS(s) +H_(2)O_(2) (aq) to `
(ii)`CO(g) + 2H_(2) (g) underset("catalyst")overset("cobalts")to`

Explanation

Let's solve the equations step by step.

### Step 1: Completing the First Equation
The first equation is:
\[ \text{PbS(s)} + \text{H}_2\text{O}_2\text{(aq)} \to \]

1. **Identify Reactants**: We have lead sulfide (PbS) and hydrogen peroxide (H₂O₂).
2. **Determine Products**: When PbS reacts with H₂O₂, it forms lead sulfate (PbSO₄) and water (H₂O).
3. **Write the Balanced Equation**:
- The unbalanced equation is:
\[ \text{PbS} + \text{H}_2\text{O}_2 \to \text{PbSO}_4 + \text{H}_2\text{O} \]
- To balance the equation, we need to ensure the number of atoms for each element is the same on both sides.
- The balanced equation is:
\[ \text{PbS} + 4 \text{H}_2\text{O}_2 \to \text{PbSO}_4 + 4 \text{H}_2\text{O} \]

### Step 2: Completing the Second Equation
The second equation is:
\[ \text{CO(g)} + 2\text{H}_2\text{(g)} \underset{\text{catalyst}}{\overset{\text{cobalt}}{\to}} \]

1. **Identify Reactants**: We have carbon monoxide (CO) and hydrogen gas (H₂).
2. **Determine Products**: In the presence of a cobalt catalyst, CO and H₂ react to form methanol (CH₃OH).
3. **Write the Balanced Equation**:
- The unbalanced equation is:
\[ \text{CO} + 2 \text{H}_2 \to \text{CH}_3\text{OH} \]
- This equation is already balanced as there is one carbon atom, four hydrogen atoms, and one oxygen atom on both sides.
- The final balanced equation is:
\[ \text{CO} + 2 \text{H}_2 \underset{\text{catalyst}}{\overset{\text{cobalt}}{\to}} \text{CH}_3\text{OH} \]

### Final Completed Equations
1. \( \text{PbS} + 4 \text{H}_2\text{O}_2 \to \text{PbSO}_4 + 4 \text{H}_2\text{O} \)
2. \( \text{CO} + 2 \text{H}_2 \underset{\text{catalyst}}{\overset{\text{cobalt}}{\to}} \text{CH}_3\text{OH} \)