The limiting reactant given that 2.2 g of Mg reacts with 4.5 L of oxygen at STP is magnesium, Mg
How to determine the mass of Oxygen
22.4 L = 1 mole of oxygen at STP
But
1 mole of oxygen = 32 g
Thus,
22.4 L = 32 g of oxygen at STP
Therefore,
4.5 L = (4.5 × 32) / 22.4
4.5 L = 6.4 g of oxygen
How to determine the limiting reactant
Balanced equation
2Mg + O₂ —> 2MgO
Molar mass of Mg = 24 g/mole
Mass of Mg from the balanced equation = 2 × 24 = 48 g
Molar mass of O₂ = 32 g/mole
Mass of O₂ from the balanced equation = 1 × 32 = 32 g
From the balanced equation above,
48 g of Mg reacted with 32 g of O₂
Therefore,
2.2 g of Mg will react with = (2.2 × 32) / 48 = 1.5 g of O₂
From the calculation made above, we can see that only 1.5 g of O₂ out of 6.5 g is needed to react completely with 2.2 g of Mg.
Thus, Mg is the limiting reactant
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