Answer:
After the reaction, there will 0.60 g of magnesium oxide and 0.25 g of oxygen gas present in the tube
Explanation:
Equation of the reaction between magnesium and oxygen is given as follows:
2Mg(s) + O₂(g) ---> 2MgO(s)
From the equation of reaction, 2 moles of magnesium reacts with i mole of oxygen gas to produce 1 mole of magnesium oxide
molar mass of magnesium is 24.0 g; molar mass of oxygen gas = 32.0 g; molar mass of magnesium oxide = 40.0 g
Therefore 24 g of magnesium reacts with 32 g of oxygen gas
I.00 g of magnesium will react with (24.0 / 32.0) * 1.00 g of oxygen = 0.75 g of oxygen gas.
Therefore, magnesium is the limiting reagent. Once it is used up, the reaction will stop and the excess oxygen will be left in the tube together with the product, magnesium oxide.
mass of excess oxygen = 1.00 - 0.75 = 0.25 g
mass of magnesium oxide formed = (24.0 / 40.0 g) * 1 = 0.60 g
