Answer :
The balanced chemical equation would be as follows:
Mg + O2 → MgO2
We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:
62.0 g Mg (1 mol / 24.31 g ) = 2.55 mol Mg
55.5 g O2 ( 1 mol / 32 g ) = 1.74 mol O2 -----> consumed completely and therefore the limiting reactant
2.55 mol - 1.74 mol O2( 1 mol Mg / 1 mol O2 ) = 0.81 mol Mg excess
Mg + O2 → MgO2
We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:
62.0 g Mg (1 mol / 24.31 g ) = 2.55 mol Mg
55.5 g O2 ( 1 mol / 32 g ) = 1.74 mol O2 -----> consumed completely and therefore the limiting reactant
2.55 mol - 1.74 mol O2( 1 mol Mg / 1 mol O2 ) = 0.81 mol Mg excess
Answer:
14.1648 grams of oxygen gas will be left.
Explanation:
[tex]2Mg +O_2\rightarrow 2MgO[/tex]
Moles of magnesium metal =[tex]\frac{62.0 g}{24 g/mol}=2.5833 mol[/tex]
Moles of oxygen gas =[tex]\frac{55.5 g}{32 g/mol}=1.7343 mol[/tex]
According to reaction, 2 mol of magnesium react with 1 mol of oxygen gas .
Then 2.5833 moles of magnesium will react with:
[tex]\frac{1}{2}\times 2.5833 mol=1.29165 mol[/tex] of oxygen gas.
Moles of oxygen left unreacted =1.7343 mol - 1.29165 mol = 0.44265 mol
Oxygen gas is an excessive reagent.
Mass of 0.44265 moles of oxygen gas:
0.44265 mol × 32 g/mol = 14.1648 g
14.1648 grams of oxygen gas will be left.