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?

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