Question

9.87 g of calcium sulfate and 12.05 g of potassium react. What is the total amount of potassium sulfate that can be produced?

Answer 1

Answer:

                     Mass of K₂SO₄  =  12.633 g

Explanation:

                    The balance chemical equation for given reaction is as;

                                     CaSO₄ + 2 K = K₂SO₄ + Ca

To solve this reaction one should first find the limiting reagent. To do so e will calculate the moles of each reactant as,

Moles of CaSO₄  =  Given Mass / M.Mass of CaSO₄

Moles of CaSO₄  =  9.87 g / 136.14 g/mol

Moles of CaSO₄  =  0.0725 moles

Similarly for K,

Moles of K  =  Given Mass / A.Mass of K

Moles of K  =  12.05 g / 39.10 g/mol

Moles of K  =  0.308 moles

Now, according to equation,

             1 mole of CaSO₄ reacts with  =  2 moles of K

So,

         0.0725 moles of CaSO₄ will react with  =  X moles of K

Solving for X,

                     X  =  0.0725 moles × 2 moles / 1 mole

                      X  =  0.145 moles of K

As calculated above, we are provided with 0.308 moles of K while, we require only 0.145 moles of it so it means that K is in excess and CaSO₄ is the limiting reagent hence, CaSO₄ will control the yield of K₂SO₄.

So,

The amount of K₂SO₄ produced is calculated by first finding its moles as,

According to equation,

             1 mole of CaSO₄ produced  =  1 mole of K₂SO₄

So,

         0.0725 moles of CaSO₄ will produce  =  X moles of K₂SO₄

Solving for X,

                     X  =  0.0725 moles × 1 mole / 1 mole

                      X  =  0.0725 moles of K₂SO₄

Now convert moles of K₂SO₄ to mass as,

Mass of K₂SO₄  =  Moles × M.Mass

Mass of K₂SO₄  =  0.0725 × 174.26 g/mol

Mass of K₂SO₄  =  12.633 g