What volume (in mililiters) of oxygen gas is required to react with 4.03 g of my at stp

You must use 1880 mL of O₂ to react with 4.03 g Mg.

A_r: 24.305

2Mg + O₂ ⟶ 2MgO

Moles of Mg = 4.03 g Mg * (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg

Moles of O₂ = 0.1658 mol Mg * (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂

STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of 22.71 L.

Volume of O₂ = 0.082 90 mol O₂ * (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL