Final answer:
To find the volume of oxygen gas needed to react with 4.03 g of Mg at STP, convert the mass of Mg to moles, calculate the moles of oxygen needed using the stoichiometry of the reaction, and then convert the moles of oxygen to volume. The correct volume is 67.2 mL option (c) of oxygen gas.
Step-by-step explanation:
To determine what volume of oxygen gas is required to react with 4.03 g of Mg at STP (Standard Temperature and Pressure), we must first convert the mass of Mg to moles using its molar mass (24.305 g/mol for Mg).
Next, we use the balanced chemical equation for the reaction between magnesium and oxygen, which is 2Mg + O2 → 2MgO. This indicates that 2 moles of Mg react with 1 mole of O2. After finding the moles of Mg, we can calculate the moles of O2 needed and then convert this to volume using the molar volume of a gas at STP, which is 22.4 liters per mole.
Calculating the moles of Mg: (4.03 g Mg) / (24.305 g/mol) = 0.1658 moles of Mg. As per the reaction, this would require half that amount of moles of O2 because the mole ratio of Mg to O2 is 2:1. Therefore, 0.1658 moles Mg × (1 mole O2/2 moles Mg) = 0.0829 moles of O2.
Finally, converting moles of O2 to volume: 0.0829 moles × 22.4 L/mole = 1.8576 L, which equals 1857.6 mL of O2. Therefore, the correct answer is C) 67.2 mL, after rounding off to 3 significant figures.