If 24.7 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?
2 NO (g) + O₂ (g) → 2 NO₂ (g

To solve this problem, we need to follow these steps:

1. **Write the balanced equation**:
   \[
   2 \text{NO (g)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{NO}_2 \text{(g)}
   \]
   
   From the balanced equation, we see that 2 moles of NO react with 1 mole of O₂.

2. **Calculate the moles of NO and O₂**:
   - Molar mass of NO: \(14.01 + 16.00 = 30.01 \, \text{g/mol}\)
   - Molar mass of O₂: \(16.00 \times 2 = 32.00 \, \text{g/mol}\)

   Now, we calculate the moles for each reactant:

   - Moles of NO: 
     \[
     \frac{24.7 \, \text{g}}{30.01 \, \text{g/mol}} = 0.823 \, \text{mol NO}
     \]
   
   - Moles of O₂:
     \[
     \frac{13.8 \, \text{g}}{32.00 \, \text{g/mol}} = 0.431 \, \text{mol O}_2
     \]

3. **Determine the limiting reactant**:
   According to the balanced equation, 2 moles of NO react with 1 mole of O₂, so the mole ratio is:
   \[
   \frac{2 \, \text{mol NO}}{1 \, \text{mol O}_2}
   \]

   To find the limiting reactant, compare the available moles of each substance based on this ratio:
   
   - For 0.823 mol of NO, the amount of O₂ required is:
     \[
     \frac{0.823 \, \text{mol NO}}{2} = 0.4115 \, \text{mol O}_2
     \]
     Since we have 0.431 mol O₂, and only 0.4115 mol is needed to react with all of the NO, NO is the limiting reactant.

4. **Calculate the excess O₂**:
   Since O₂ is in excess, subtract the amount of O₂ that reacted:
   \[
   0.431 \, \text{mol O}_2 - 0.4115 \, \text{mol O}_2 = 0.0195 \, \text{mol O}_2
   \]

So, **0.0195 moles of O₂** will be left over as the excess reactant.

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Would you like more details, or do you have any questions? Here are some related questions to consider:

1. How do you identify the limiting reactant in a reaction?
2. What are the steps to calculate the molar mass of a compound?
3. How does the mole ratio from the balanced equation help in limiting reactant problems?
4. Why is it important to identify the limiting reactant in chemical reactions?
5. How can the concept of limiting reactants be applied in industrial chemical processes?

**Tip:** Always balance the chemical equation first when solving stoichiometry problems—it ensures correct mole ratios!

If 24.7 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over? 2 NO (g) + O₂ (g) → 2 NO₂ (g)

Learn how to determine the limiting reactant and calculate the moles of excess O₂ left over in the reaction between NO and O₂. This detailed solution uses stoichiometry, the mole concept, and balanced chemical equations. Suitable for students in Grades 10-12.

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