How many grams of Al will react with 28 L O2 at STP?

Answer 1

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, 28 L of O₂ corresponds to ( \frac{28}{22.4} \approx 1.25 ) moles of O₂. The balanced reaction between aluminum (Al) and oxygen (O₂) is ( 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 ), indicating that 4 moles of Al react with 3 moles of O₂. Thus, for 1.25 moles of O₂, ( \frac{4}{3} \times 1.25 \approx 1.67 ) moles of Al are required, which corresponds to approximately ( 1.67 \times 27 \approx 45 ) grams of Al.