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

What is the maximum amount of water (in grams) that can be produced when 964.91 g of C3H8 reacts with 511.39 g of oxygen according to the balanced equation: C3H8 + 5 O2 -> 3 CO2 + 4 H2O?

327.68 g

400.00 g

600.00 g

500.00 g

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Verified step by step guidance

First, identify the balanced chemical equation for the reaction: \( \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \). This equation shows the stoichiometry of the reactants and products.

Calculate the molar mass of \( \text{C}_3\text{H}_8 \) (propane) and \( \text{O}_2 \) (oxygen). The molar mass of \( \text{C}_3\text{H}_8 \) is \( 3 \times 12.01 + 8 \times 1.01 = 44.11 \text{ g/mol} \). The molar mass of \( \text{O}_2 \) is \( 2 \times 16.00 = 32.00 \text{ g/mol} \).

Convert the mass of \( \text{C}_3\text{H}_8 \) and \( \text{O}_2 \) to moles using their respective molar masses. For \( \text{C}_3\text{H}_8 \), use \( \frac{964.91 \text{ g}}{44.11 \text{ g/mol}} \). For \( \text{O}_2 \), use \( \frac{511.39 \text{ g}}{32.00 \text{ g/mol}} \).

Determine the limiting reactant by comparing the mole ratio from the balanced equation. The balanced equation requires 1 mole of \( \text{C}_3\text{H}_8 \) to react with 5 moles of \( \text{O}_2 \). Calculate the required moles of \( \text{O}_2 \) for the available moles of \( \text{C}_3\text{H}_8 \) and compare it with the available moles of \( \text{O}_2 \).

Using the limiting reactant, calculate the maximum moles of \( \text{H}_2\text{O} \) produced. From the balanced equation, 1 mole of \( \text{C}_3\text{H}_8 \) produces 4 moles of \( \text{H}_2\text{O} \). Convert the moles of \( \text{H}_2\text{O} \) to grams using the molar mass of water (\( 18.02 \text{ g/mol} \)).