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In the question shown in the image, calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. You are tasked with finding the mass of each substance after 23.24 g of calcium nitrate and 24.22 g of ammonium fluoride react completely.

To solve this, you need to use stoichiometry, based on the balanced chemical equation for the reaction. Here are the general steps:

1. **Write the balanced chemical equation** for the reaction:
   \[
   \text{Ca(NO}_3\text{)}_2 + 6\text{NH}_4\text{F} \rightarrow \text{CaF}_2 + 2\text{N}_2\text{O} + 4\text{H}_2\text{O}
   \]
   This equation tells us the molar ratios between the reactants and products.

2. **Find the molar masses** of the reactants:
   - **Calcium nitrate (Ca(NO₃)₂)**: 
     \[
     \text{Ca} = 40.08 \, \text{g/mol}, \, \text{N} = 14.01 \, \text{g/mol}, \, \text{O} = 16.00 \, \text{g/mol}
     \]
     So, the molar mass of calcium nitrate is:
     \[
     40.08 + 2(14.01 + 3(16.00)) = 164.1 \, \text{g/mol}
     \]
   - **Ammonium fluoride (NH₄F)**: 
     \[
     \text{N} = 14.01 \, \text{g/mol}, \, \text{H} = 1.008 \, \text{g/mol}, \, \text{F} = 19.00 \, \text{g/mol}
     \]
     So, the molar mass of ammonium fluoride is:
     \[
     14.01 + 4(1.008) + 19.00 = 37.05 \, \text{g/mol}
     \]

3. **Convert the given masses** of calcium nitrate and ammonium fluoride into moles:
   - Moles of calcium nitrate:
     \[
     \text{moles of Ca(NO}_3\text{)}_2 = \frac{23.24 \, \text{g}}{164.1 \, \text{g/mol}} = 0.1416 \, \text{mol}
     \]
   - Moles of ammonium fluoride:
     \[
     \text{moles of NH}_4\text{F} = \frac{24.22 \, \text{g}}{37.05 \, \text{g/mol}} = 0.6536 \, \text{mol}
     \]

4. **Determine the limiting reactant**:
   According to the balanced equation, 1 mole of calcium nitrate reacts with 6 moles of ammonium fluoride. So:
   \[
   \frac{0.1416 \, \text{mol of Ca(NO}_3\text{)}_2}{1} \text{ requires } 0.1416 \times 6 = 0.8496 \, \text{mol of NH}_4\text{F}
   \]
   However, we only have 0.6536 mol of ammonium fluoride, so ammonium fluoride is the limiting reactant.

5. **Calculate the masses of products**:
   Using the limiting reactant (NH₄F), you can calculate the moles of each product and convert them into grams.

Would you like me to continue with the calculations of the masses of the products, or do you have any specific questions about these steps?

---

Here are 5 follow-up questions to consider:
1. How is the concept of limiting reactants important in chemical reactions?
2. Can you identify other examples of real-life applications of stoichiometry?
3. What happens when there is an excess reactant in a reaction?
4. Why is it important to use significant figures in chemical calculations?
5. How can we experimentally determine the products of this reaction?

**Tip**: Always ensure the chemical equation is balanced before starting any stoichiometry problem, as the molar ratios are key to solving the problem.

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. What mass of each substance is present after 23.24 g of calcium nitrate and 24.22 g of ammonium fluoride react completely?

Learn how to calculate the masses of products formed in a stoichiometric reaction involving calcium nitrate and ammonium fluoride. This problem covers stoichiometric concepts such as limiting reactant and mass calculations, with a balanced chemical equation. Suitable for high school students in Grades 10-12.

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