Final answer:
Using stoichiometry and the balanced chemical equation, 24.4 g of O₂ can produce 77.26 g of chromium(III) oxide (Cr₂O₃) when completely reacted.
Step-by-step explanation:
To calculate how many grams of chromium(III) oxide (Cr₂O₃) form when 24.4 g of O₂ completely reacts, we use stoichiometry based on the balanced chemical equation 4Cr(s) + 3O₂(g) → 2Cr₂O₃(s). First, we calculate the number of moles of O₂ using the molar mass of O₂, which is approximately 32 g/mol. Since 24.4 g of O₂ divide by 32 g/mol gives us 0.7625 moles of O₂.
From the chemical equation, the mole ratio of O₂ to Cr₂O₃ is 3:2, which implies 0.7625 moles of O₂ will produce (0.7625 moles × 2 mol Cr₂O₃ / 3 mol O₂) = 0.5083 moles of Cr₂O₃. To find the mass of Cr₂O₃ produced, we multiply the moles of Cr₂O₃ by the molar mass of Cr₂O₃ (approximately 152 g/mol), giving us (0.5083 moles × 152 g/mol) = 77.26 g of Cr₂O₃. Therefore, when 24.4 g of O₂ completely react, 77.26 g of Cr₂O₃ form.