When sulfuric acid reacts with zinc hydroxide, zinc sulfate and water are produced. The balanced equation for this reaction is: H2 SO4 (aq) + Zn(OH)2(s) → ZnSO4 (aq) + 2H2O(1) If 4 moles of sulfurio

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**Chemical Reaction of Sulfuric Acid and Zinc Hydroxide**

When sulfuric acid reacts with zinc hydroxide, zinc sulfate and water are produced. The balanced equation for this reaction is:

\[ H_2SO_4(aq) + Zn(OH)_2(s) \rightarrow ZnSO_4(aq) + 2H_2O(l) \]

If 4 moles of sulfuric acid react:

- The reaction consumes \[ \boxed{4} \] moles of zinc hydroxide.
- The reaction produces \[ \boxed{4} \] moles of zinc sulfate and \[ \boxed{8} \] moles of water.
Transcribed Image Text:**Chemical Reaction of Sulfuric Acid and Zinc Hydroxide** When sulfuric acid reacts with zinc hydroxide, zinc sulfate and water are produced. The balanced equation for this reaction is: \[ H_2SO_4(aq) + Zn(OH)_2(s) \rightarrow ZnSO_4(aq) + 2H_2O(l) \] If 4 moles of sulfuric acid react: - The reaction consumes \[ \boxed{4} \] moles of zinc hydroxide. - The reaction produces \[ \boxed{4} \] moles of zinc sulfate and \[ \boxed{8} \] moles of water.
**Chemical Reaction: Ammonia and Oxygen**

When ammonia reacts with oxygen, nitrogen monoxide and water are produced. The balanced equation for this reaction is:

\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \]

**Problem Statement:**

Given: 15 moles of oxygen react,

**Questions:**

1. The reaction consumes \_\_\_\_ moles of ammonia.
2. The reaction produces \_\_\_\_ moles of nitrogen monoxide and \_\_\_\_ moles of water.

**Instructions for Students:**

1. Use the stoichiometric coefficients from the balanced equation to find the moles of ammonia, nitrogen monoxide, and water involved.
2. Apply the concept of stoichiometry to convert moles of oxygen to moles of each reactant and product based on the balanced equation.

**Example Calculation:**

Given the balanced equation:

\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \]

- **Step 1:** Identify mole ratio from the equation:
  - 5 moles of \( \text{O}_2 \) reacts with 4 moles of \( \text{NH}_3 \).
  - 5 moles of \( \text{O}_2 \) produce 4 moles of \( \text{NO} \).
  - 5 moles of \( \text{O}_2 \) produce 6 moles of \( \text{H}_2\text{O} \).

- **Step 2:** Calculate the moles using given data (15 moles of \( \text{O}_2 \)):

\[ \text{Moles of NH}_3 = \left( \frac{4 \text{ moles NH}_3}{5 \text{ moles O}_2} \right) \times 15 \text{ moles O}_2 \]
\[ = 12 \text{ moles NH}_3 \]

\[ \text{Moles of NO} = \left( \frac{4 \text{ moles NO}}{5 \
Transcribed Image Text:**Chemical Reaction: Ammonia and Oxygen** When ammonia reacts with oxygen, nitrogen monoxide and water are produced. The balanced equation for this reaction is: \[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \] **Problem Statement:** Given: 15 moles of oxygen react, **Questions:** 1. The reaction consumes \_\_\_\_ moles of ammonia. 2. The reaction produces \_\_\_\_ moles of nitrogen monoxide and \_\_\_\_ moles of water. **Instructions for Students:** 1. Use the stoichiometric coefficients from the balanced equation to find the moles of ammonia, nitrogen monoxide, and water involved. 2. Apply the concept of stoichiometry to convert moles of oxygen to moles of each reactant and product based on the balanced equation. **Example Calculation:** Given the balanced equation: \[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \] - **Step 1:** Identify mole ratio from the equation: - 5 moles of \( \text{O}_2 \) reacts with 4 moles of \( \text{NH}_3 \). - 5 moles of \( \text{O}_2 \) produce 4 moles of \( \text{NO} \). - 5 moles of \( \text{O}_2 \) produce 6 moles of \( \text{H}_2\text{O} \). - **Step 2:** Calculate the moles using given data (15 moles of \( \text{O}_2 \)): \[ \text{Moles of NH}_3 = \left( \frac{4 \text{ moles NH}_3}{5 \text{ moles O}_2} \right) \times 15 \text{ moles O}_2 \] \[ = 12 \text{ moles NH}_3 \] \[ \text{Moles of NO} = \left( \frac{4 \text{ moles NO}}{5 \
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