11.775 g of magnesium is reacted with 4.938 g of oxygen gas according to the equation: 2 Mg + O2 --> 2 MgO, what is the limiting reactant?

Magnesium is clearly the reagent in excess according to the reaction stoichiometry; a stoichiometric reaction would require 0.968 mol of oxygen gas.

To determine the limiting reactant, we need to compare the amounts of magnesium and oxygen gas present and see which one is consumed completely in the reaction.

First, we need to convert the masses of magnesium and oxygen gas to moles using their respective molar masses.

• The molar mass of magnesium (Mg) is approximately 24.305 g/mol.
• The molar mass of oxygen (O2) is approximately 32.00 g/mol.

Next, we calculate the number of moles of each reactant:

• Moles of magnesium = Mass of magnesium / Molar mass of magnesium
• Moles of oxygen = Mass of oxygen / Molar mass of oxygen

Once we have the moles of each reactant, we look at the stoichiometric coefficients in the balanced equation to see the ratio in which they react. In this case, 2 moles of magnesium react with 1 mole of oxygen.

After finding the moles of each reactant, we can compare the stoichiometric ratios to determine the limiting reactant. The reactant that is present in the lesser amount, relative to the stoichiometric ratio, will be the limiting reactant.