Math Problem Statement
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
Solution
To solve this problem, we need to follow these steps:
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Write the balanced equation:
From the balanced equation, we see that 2 moles of NO react with 1 mole of O₂.
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Calculate the moles of NO and O₂:
- Molar mass of NO:
- Molar mass of O₂:
Now, we calculate the moles for each reactant:
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Moles of NO:
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Moles of O₂:
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Determine the limiting reactant: According to the balanced equation, 2 moles of NO react with 1 mole of O₂, so the mole ratio is:
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: 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.
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Calculate the excess O₂: Since O₂ is in excess, subtract the amount of O₂ that reacted:
So, 0.0195 moles of O₂ will be left over as the excess reactant.
Would you like more details, or do you have any questions? Here are some related questions to consider:
- How do you identify the limiting reactant in a reaction?
- What are the steps to calculate the molar mass of a compound?
- How does the mole ratio from the balanced equation help in limiting reactant problems?
- Why is it important to identify the limiting reactant in chemical reactions?
- 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!
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Math Problem Analysis
Mathematical Concepts
Stoichiometry
Mole Concept
Limiting Reactant
Excess Reactant
Formulas
Moles = mass / molar mass
Stoichiometric ratio: 2 mol NO : 1 mol O₂
Theorems
Law of Conservation of Mass
Stoichiometric Ratios in Balanced Equations
Suitable Grade Level
Grades 10-12