Assuming an efficiency of 34.90%, calculate the actual yield of magnesium nitrate formed from 139.6 g of magnesium and excess copper(II) nitrate. Mg+Cu(NO3)2⟶Mg(NO3)2+Cu

300.44 g

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

the link still does not work even when adding it to the url bar

explanation: