How many grams of magnesium oxide could be formed from the reaction of .248g of magnesium metal?

Mg--->MgO

.248 g Mg x (1 mol Mg/MM Mg) x (1 mol MgO/1 mol Mg) x (MM MgO/1 mol MgO)

MM are molar masses.

To determine the number of grams of magnesium oxide that could be formed, we need to consider the balanced chemical equation for the reaction of magnesium metal (Mg) with oxygen (O₂) to form magnesium oxide (MgO):

2Mg + O₂ → 2MgO

From the balanced equation, we can see that 2 moles of magnesium (Mg) react with 1 mole of oxygen (O₂) to produce 2 moles of magnesium oxide (MgO).

To get the molar mass of magnesium (Mg), we look at the periodic table and find that it is approximately 24.31 g/mol.

Now, we can calculate the number of moles of magnesium metal using the given mass:

moles of magnesium (Mg) = mass of magnesium (Mg) / molar mass of magnesium (Mg)
= 0.248 g / 24.31 g/mol

Next, we use the mole ratio from the balanced equation to determine the number of moles of magnesium oxide (MgO) formed:

moles of magnesium oxide (MgO) = moles of magnesium (Mg) × (2 moles of MgO / 2 moles of Mg)

Finally, we can convert the moles of magnesium oxide (MgO) to grams:

mass of magnesium oxide (MgO) = moles of magnesium oxide (MgO) × molar mass of magnesium oxide (MgO)

Now, let's substitute the values into the equations:

moles of magnesium (Mg) = 0.248 g / 24.31 g/mol ≈ 0.01019 mol
moles of magnesium oxide (MgO) = 0.01019 mol × (2 mol MgO / 2 mol Mg) = 0.01019 mol
mass of magnesium oxide (MgO) = 0.01019 mol × (40.30 g/mol) ≈ 0.4116 g

Therefore, approximately 0.4116 grams of magnesium oxide could be formed from the reaction of 0.248 grams of magnesium metal.