Calculate the percentage yield and theoretical yield of BaSO4. Ba(NO3)2 + Na2SO4 -> BaSO4 + 2NaNO3 An experiment was done with 75.00 g of...

The theoretical yield is the maximum amount of product that can be formed in a chemical reaction. It is calculated using stoichiometry.

The actual yield is the actual amount of product formed experimentally. 

The balanced equation for this problem is:

`~Ba(NO_3)_2 + ~Na_2SO_4 -gt ~BaSO_4 +~2NaNO_3`

Calculation of the Theoretical Yield:

This is a stoichiometry calculation. The stoichiometry calculation will take the general form of:

(given substance) x (mole conversion factor) x (mole ratio)...

The theoretical yield is the maximum amount of product that can be formed in a chemical reaction. It is calculated using stoichiometry.

The actual yield is the actual amount of product formed experimentally. 

The balanced equation for this problem is:

`~Ba(NO_3)_2 + ~Na_2SO_4 -gt ~BaSO_4 +~2NaNO_3`

Calculation of the Theoretical Yield:

This is a stoichiometry calculation. The stoichiometry calculation will take the general form of:

(given substance) x (mole conversion factor) x (mole ratio) x (mole conversion factor)

Step 1: Choose the mole conversion factors.

The mole conversion factors used in stoichiometry calculations are:

• 1 mol = 22.4 L gas


• 1 mol = 6.02 x `~10^23` atoms or molecules


• 1 mol = molar mass in grams

Since both the given substance (`~Ba(NO_3)_2` ) and the final substance (`~BaSO_4` ) are both in grams, we will use the conversion factor, "1 mol = molar mass" for both conversion factors indicated in the calculation.

Step 2: Calculate the molar masses needed for the conversion factors.

The molar mass of a substance is calculated by multiplying the atomic mass of each atom times its subscript and adding the resulting answers together. 

molar mass of `~Ba(NO_3)_2` :

   [1 x 137.318] + [2 x 14.007] + [6 x 15.999] = 261.336 g

So, the first mole conversion factor in our calculation will be: 1 mol = 261.336 g `~Ba(NO_3)_2`

molar mass of `~BaSO_4` :

   [1 x 137.318] + [1 x 32.066] + [4 x 15.999] = 233.39 g

So, the second mole conversion factor in our calculation will be: 1 mol = 233.39 g `~BaSO_4`

Step 3: Perform the stoichiometry calculation.

Recall that the general form of the stoichiometry calculation is:

(given substance) x (mole conversion factor) x (mole ratio) x (mole conversion factor)

•    The given substance is found in the problem: 75.00 g `~Ba(NO_3)_2`


•    The first mole conversion factor is: 1 mol = 261.336 g `~Ba(NO_3)_2`


•    The mole ratio is: 1 mol `~BaSO_4` to 1 mol `~Ba(NO_3)_2`

•    The second mole conversion factor is: 1 mol = 233.39 g `~BaSO_4`

Plug in the values and calculate:

(75.00 g)(1mol/261.336 g)(1mol/1 mol)(233.39 g/1 mol) = 66.98 g` <br> `

Therefore, the theoretical yield = 66.98 g `~BaSO_4`

*Notice that the conversion factors in the calculation are oriented such that the unit in the denominator is the same as the unit in the numerator in the previous section of the calculation - this ensures that units cancel out leaving only the final unit.

Calculation of Percent Yield

The percent yield is calculated using the formula:

   % yield = (actual yield/theoretical yield) x 100

Therefore, 

   % yield = (63.45 g/66.98 g) x 100 = 94.73%