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 volume of ammonia, NH3, in liters at STP can be produced from 2.0 kg of hydrogen gas, H2, and an excess of nitrogen gas, N2? (N2 (g) + 3 H2 (g) → 2 NH3 (g))

74.6 L

22.4 L

149.2 L

44.8 L

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Verified step by step guidance

First, convert the mass of hydrogen gas (H2) from kilograms to grams. Since 1 kg = 1000 g, multiply 2.0 kg by 1000 to get 2000 g of H2.

Next, calculate the number of moles of hydrogen gas using its molar mass. The molar mass of H2 is approximately 2.02 g/mol. Use the formula: \( \text{moles of H2} = \frac{\text{mass of H2}}{\text{molar mass of H2}} \).

Using the balanced chemical equation \( \text{N}_2 (g) + 3 \text{H}_2 (g) \rightarrow 2 \text{NH}_3 (g) \), determine the mole ratio between H2 and NH3. From the equation, 3 moles of H2 produce 2 moles of NH3.

Calculate the moles of NH3 produced using the mole ratio. Use the formula: \( \text{moles of NH3} = \frac{2}{3} \times \text{moles of H2} \).

Finally, convert the moles of NH3 to volume at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies 22.4 liters. Use the formula: \( \text{volume of NH3} = \text{moles of NH3} \times 22.4 \text{ L/mol} \).