Answer:
300.44 g
Step-by-step explanation:
The balanced equation for the reaction is given below:
Mg + Cu(NO3)2 —> Mg(NO3)2 + Cu
Next, we shall determine the mass of Mg that reacted and the mass of Mg(NO3)2 produced from the balanced equation.
This is illustrated below:
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 x 24 = 24 g
Molar mass of Mg(NO3)2 = 24 + 2[14 + (16x3)]
= 24 + 2[ 14 + 48]
= 24 + 124 = 148 g/mol
Mass of Mg(NO3)2 from the balanced equation =
1 x 148 = 148 g
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Next, we shall determine the theoretical yield of Mg(NO3)2.
This can be obtained as follow:
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Therefore, 139.6 g of Mg will react to = (139.6 x 148)/24 = 860.87 g of Mg(NO3)2
Therefore, the theoretical yield of Mg(NO3)2 is 860.87 g
Finally, we shall determine the actual yield of Mg(NO3)2 as follow:
Theoretical of Mg(NO3)2 = 860.87 g
Percentage yield = 34.90%
Actual yield of Mg(NO3)2 =?
Percentage yield = Actual yield /Theoretical yield x 100
34.90% = Actual yield /860.87
Cross multiply
Actual yield = 34.90% x 860.87
Actual yield = 34.9/100 x 860.87
Actual yield = 300.44 g
Therefore, the actual yield of Mg(NO3)2 is 300.44 g