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)
Added by Loretta M.
Step 1
The molar mass of NO is approximately 30 g/mol (14 g/mol for N and 16 g/mol for O). So, 24.7 g of NO is 24.7 g / 30 g/mol = 0.823 moles. The molar mass of O₂ is approximately 32 g/mol (16 g/mol for each O). So, 13.8 g of O₂ is 13.8 g / 32 g/mol = 0.431 Show more…
Show all steps
Your feedback will help us improve your experience
Suman K and 91 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If 23.4 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)
Sima S.
If 22.1 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)
Supreeta N.
Assume that you have $1.39 \mathrm{~mol}$ of $\mathrm{H}_{2}$ and $3.44 \mathrm{~mol}$ of $\mathrm{N}_{2}$. How many grams of ammonia $\left(\mathrm{NH}_{3}\right)$ can you make, and how many grams of which reactant will be left over? $$ 3 \mathrm{H}_{2}+\mathrm{N}_{2} \longrightarrow 2 \mathrm{NH}_{3} $$
Mass Relationships in Chemical Reactions
Determining Molecular Weights: Mass Spectrometry
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD