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


$4Cr\left(s\right)+3O_2\left(g\right)⟶2C{r}_2{O}_3\left(s\right)$


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

• A. 0.334 mol
• B. 1.33 mol
• C. 0.749 mol
• D. 3.00 mol
• E. 0.667 mol

Answer: (A)

Verified step by step guidance

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.