in reaction = Mg + O2 --> MgO
if 24 gram of Mg is burned in excess oxygen to form 40 gram of MgO, How much O2 is used?
Refer to your last post on SO2
To find out how much O2 is used in the reaction, we need to first determine the amount of MgO that can be produced from the given amount of Mg.
1. Find the molar mass of Mg:
The molar mass of Mg is 24.31 g/mol.
2. Convert the given mass of Mg to moles:
Mass of Mg = 24 g
Moles of Mg = (mass of Mg) / (molar mass of Mg) = 24 g / 24.31 g/mol ≈ 0.987 mol
3. Use the balanced chemical equation to determine the mole ratio of Mg to MgO:
According to the balanced equation: 2 moles of Mg reacts with 1 mole of O2 to produce 2 moles of MgO.
4. Determine the moles of MgO that can be produced:
Moles of MgO = (moles of Mg) / 2 = 0.987 mol / 2 = 0.494 mol
5. Find the molar mass of MgO:
The molar mass of MgO is 40.31 g/mol.
6. Convert the moles of MgO to mass:
Mass of MgO = (moles of MgO) * (molar mass of MgO) = 0.494 mol * 40.31 g/mol ≈ 19.92 g
Therefore, when 24 grams of Mg is burned in excess oxygen to form 40 grams of MgO, the amount of O2 used can be calculated by subtracting the mass of MgO produced from the initial mass of Mg:
Mass of O2 used = mass of Mg - mass of MgO = 24 g - 19.92 g ≈ 4.08 g
So, approximately 4.08 grams of O2 is used in the reaction.