Table of contents

1. The Chemical World9m
- The Scientific Method
  9m
2. Measurement and Problem Solving2h 19m
3. Matter and Energy2h 15m
4. Atoms and Elements2h 33m
5. Molecules and Compounds1h 50m
6. Chemical Composition1h 23m
7. Chemical Reactions1h 43m
8. Quantities in Chemical Reactions1h 8m
9. Electrons in Atoms and the Periodic Table2h 32m
10. Chemical Bonding2h 10m
11 Gases2h 7m
12. Liquids, Solids, and Intermolecular Forces1h 11m
13. Solutions3h 1m
14. Acids and Bases2h 14m
15. Chemical Equilibrium1h 27m
16. Oxidation and Reduction1h 33m
17. Radioactivity and Nuclear Chemistry53m

8. Quantities in Chemical Reactions

Stoichiometry

Introduction to Chemistry

8. Quantities in Chemical Reactions

Stoichiometry

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Multiple Choice

The oxidation of chromium solid is represented by the following equation:

How many moles of chromium (III) oxide are produced when 34.69 g Cr reacts with excess oxygen gas?

0.334 mol

1.33 mol

0.749 mol

3.00 mol

0.667 mol

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Verified step by step guidance

First, identify the balanced chemical equation for the reaction: 4 Cr(s) + 3 O₂(g) → 2 Cr₂O₃(s). This equation shows that 4 moles of chromium react with 3 moles of oxygen to produce 2 moles of chromium(III) oxide.

Calculate the molar mass of chromium (Cr). The atomic mass of chromium is approximately 51.996 g/mol. Therefore, the molar mass of Cr is 51.996 g/mol.

Determine the number of moles of chromium in 34.69 g. Use the formula: moles = mass (g) / molar mass (g/mol). Substitute the given mass of chromium and its molar mass into the formula to find the moles of Cr.

Use the stoichiometry of the balanced equation to find the moles of chromium(III) oxide produced. According to the equation, 4 moles of Cr produce 2 moles of Cr₂O₃. Set up a ratio using the moles of Cr calculated in the previous step to find the moles of Cr₂O₃.

Finally, compare the calculated moles of Cr₂O₃ with the given options (0.334 mol, 1.33 mol, 0.749 mol, 3.00 mol, 0.667 mol) to determine which one matches your calculation.