Answer :
Magnesium, Mg, is the limiting reactant when 2.2 g of Mg react with 4.5 L of oxygen at STP.
To determine the mass of oxygen.
1 mole of oxygen at STP = 22.4 L
But, we have
1 mole of oxygen at STP = 32g
Therefore,
22.4 L = 32 g of oxygen at STP.
It is given that there is 4.5L of oxygen is reacted with the magnesium.
So,
4.5 L = (4.5 × 32) / 22.4
4.5 L = 6.4 g of oxygen.
2Mg + O₂ —> 2MgO
Molar mass of Mg = 24 g [tex]mol^{-1}[/tex]
Molar mass of O₂ = 32 g [tex]mol^{-1}[/tex]
Mass of O₂ from the equation = 1 × 32 = 32 g
To find the limiting reactant,
From the equation above,
48 g of Mg reacts with 32 g of O₂
2.2 g of Mg will react with = ?
(2.2 × 32) / 48 = 1.5 g of O₂
2.2 g of Mg will react with 1.5 g of O₂
According to the calculations above, just 1.5 g of the 6.5 g of O2 are required for 2.2 g of Mg to react.
Therefore, the limiting reactant is Magnesium(Mg).
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Complete Question:
2Mg + O₂ —> 2MgO
What is the limiting reactant if 2. 2 g of mg is reacted with 4. 5 l of oxygen at STP?
