Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

3. Chemical Reactions

Stoichiometry

General Chemistry

3. Chemical Reactions

Stoichiometry

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Multiple Choice

In the reaction 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g), if 0.300 mol of octane (C8H18) is allowed to react with 0.870 mol of oxygen (O2) and oxygen is the limiting reactant, resulting in the production of 0.626 mol of water (H2O), how much octane is left unreacted?

0.174 mol

0.100 mol

0.200 mol

0.126 mol

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Verified step by step guidance

First, identify the balanced chemical equation for the reaction: 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g). This equation tells us the stoichiometric ratios between the reactants and products.

Determine the stoichiometric ratio between octane (C8H18) and oxygen (O2) from the balanced equation. For every 2 moles of C8H18, 25 moles of O2 are required. Therefore, the ratio is 2:25.

Calculate the amount of octane that reacts with the given amount of oxygen. Since oxygen is the limiting reactant, use the stoichiometric ratio to find out how much octane reacts with 0.870 mol of O2. Use the formula: \( \text{moles of C8H18 reacted} = \frac{2}{25} \times \text{moles of O2} \).

Subtract the moles of octane that reacted from the initial moles of octane to find the moles of octane left unreacted. Use the formula: \( \text{moles of C8H18 unreacted} = \text{initial moles of C8H18} - \text{moles of C8H18 reacted} \).

Verify the calculation by checking the stoichiometry of the reaction and ensuring that the amount of water produced (0.626 mol) aligns with the stoichiometric predictions based on the limiting reactant.