Magnesium reacts with oxygen to form magnesium oxide (MgO) according to the following balanced equation: 2Mg+O2arrow 2MgO How many liters of oxygen gas (O_(2)) are required to react with 4.00 moles of magnesium (Mg) at STP? 51.3 L 45.7 L 44.8 L 50.01

To determine the volume of oxygen gas required to react with 4.00 moles of magnesium at STP, we need to use the stoichiometry of the balanced chemical equation and the ideal gas law.<br /><br />Given:<br />- Balanced equation: \(2Mg + O_2 \rightarrow 2MgO\)<br />- Moles of magnesium (Mg): 4.00 moles<br /><br />From the balanced equation, we see that 2 moles of magnesium react with 1 mole of oxygen gas.<br /><br />First, calculate the moles of oxygen gas required:<br />\[ \text{Moles of } O_2 = \frac{\text{Moles of } Mg}{2} = \frac{4.00 \text{ moles}}{2} = 2.00 \text{ moles} \]<br /><br />At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters.<br /><br />Now, calculate the volume of oxygen gas required:<br />\[ \text{Volume of } O_2 = \text{Moles of } O_2 \times 22.4 \text{ L/mol} = 2.00 \text{ moles} \times 22.4 \text{ L/mol} = 44.8 \text{ L} \]<br /><br />Therefore, the volume of oxygen gas required to react with 4.00 moles of magnesium at STP is 44.8 liters.<br /><br />The correct answer is:<br />44.8 L