Answer:
A. 2Mg(s) + O2(g) —> 2MgO(s)
B. Mg is the limiting reactant.
C. Theoretical yield of MgO is 16.83g.
D. The percentage yield is 70.7%
Explanation:
A. The balanced equation for the reaction. This is given below:
2Mg(s) + O2(g) —> 2MgO(s)
B. Determination of the limiting reactant.
First, we shall determine the mass of Mg and O2 that reacted and the mass of MgO produced from the balanced equation. This is illustrated below:
Molar mass of Mg = 24g/mol
Mass of Mg from the balanced equation = 2 x 24 = 48g.
Molar mass of O2 = 16x2 = 32g/mol.
Mass of O2 from the balanced equation = 1 x 32 = 32g
Molar mass of MgO = 24 + 16 = 40g/mol
Mass of MgO from the balanced equation = 2 x 40 = 80g
Summary:
From the balanced equation above,
48g of Mg reacted with 32g of O2 to produce 80g of MgO.
Now, we can obtain the limiting reactant as follow:
From the balanced equation above,
48g of Mg reacted with 32g of O2.
Therefore, 10.1g of Mg will react with = (10.1 x 32)/48 = 6.73g of O2.
From the calculations made above, we can see that only 6.73g out of 10.5g of O2 given is needed to react completely with 10.1g of Mg.
Therefore, Mg is the limiting reactant and O2 is the excess reactant.
C. Determination of the theoretical yield of MgO.
The limiting reactant is used in this case as it will produce the maximum yield of the reaction since all of I is used up in the reaction.
The theoretical yield can be obtain as illustrated below:
From the balanced equation above,
48g of Mg reacted to produce 80g of MgO.
Therefore, 10.1g of Mg will react to produce = (10.1 x 80)/48 = 16.83g of MgO.
Therefore, the theoretical yield of MgO is 16.83g.
D. Determination of the percentage yield.
This is illustrated below:
Actual yield of MgO = 11.9g
Theoretical yield of MgO = 16.83g
Percentage yield =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 11.9/16.83 x 100
Percentage yield = 70.7%
