If 2.4 g of magnesium metal were completely exposed to air, what would be the weight loss or gain of magnesium metal?

The question seems to be based on the reaction of magnesium with oxygen in the air to form magnesium oxide. However, in this reaction, magnesium doesn't lose weight but actually gains weight. This is because it is combining with oxygen from the air.

Here's a step-by-step explanation:

1. Write down the balanced chemical equation for the reaction: 2Mg + O2 -> 2MgO

2. Calculate the molar mass of magnesium (Mg), oxygen (O2), and magnesium oxide (MgO). Using the periodic table, we find that the molar mass of Mg is approximately 24.3 g/mol, O2 is approximately 32 g/mol, and MgO is approximately 40.3 g/mol.

3. From the balanced chemical equation, we can see that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. This means that 48.6 g of Mg (2 moles * 24.3 g/mol) react with 32 g of O2 to produce 80.6 g of MgO (2 moles * 40.3 g/mol).

4. Now, we can set up a proportion to find out how much MgO is produced when only 2.4 g of Mg reacts. If 48.6 g of Mg produces 80.6 g of MgO, then 2.4 g of Mg will produce (2.4 g * 80.6 g) / 48.6 g = 5 g of MgO (approximately).

5. Therefore, the weight gain of the magnesium metal when it is completely exposed to air is approximately 5 g - 2.4 g = 2.6 g.

So, the magnesium metal would gain approximately 2.6 g when completely exposed to air.