For the following reaction, 53.0 grams of potassium hydrogen sulfate are allowed to react with 20.0 grams of potassium hydroxide. potassium hydrogen sulfate (aq) + potassium hydroxide (aq) potassium sulfate (aq) + water (l) What is the maximum amount of potassium sulfate that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

See explanation below

Explanation:

First, in order to solve this problem we need to write the formula for each reactant and product here.

The potassium hydrogen sulfate is a salt, and this is commonly known as potassium bisulfate. If the sulfate is SO₄²⁻, then, the hydrogen sulfate would be HSO₄⁻. So potassium hydrogen sulfate would be KHSO₄.

Potassium hydroxide is a base, and it's formula is KOH.

The products of reaction would be K₂SO₄ (Potassium sulfate) and water H₂O.

Basing on this, the overall reaction is:

KHSO₄ + KOH > K₂SO₄ + H₂O

We can see that this reaction is already balanced, and we have a mole ratio of each reactant and product of 1:1, therefore, we can assume that 1 mole of bisulfate needs to react with 1 mole of KOH to produce 1 mole of each product there. So in order to know which is the limiting reactant, we just need to calculate the number of moles of each reactant basing on their masses, the one with the lowest moles would be the limiting reactant.

To calculate the moles, we need the molecular weight of the reactants which are:

MW KHSO₄ = 136.17 g/mol

MW KOH = 56.11 g/mol

MW K₂SO₄ = 174.26 g/mol

Then, the moles are:

moles KHSO₄ = 53 / 136.17 = 0.39 moles

moles KOH = 20 / 56.11 = 0.36 moles

As the moles of KOH < moles KHSO₄ we can say that the limiting reactant is the potassium hydroxide KOH

To get the amount of the excess reactant, we just see how many moles of this reacted with the KOH:

remanent moles of KHSO₄ = 0.39 - 0.36 = 0.03 moles

With this, we calculate the mass:

m = 0.03 * 136.17

m KHSO₄ = 4.09 g

Finally to get the maximum amount of potassium sulfate formed, we already know how many moles reacted, and these, are the same moles produced according to our innitial mole ratio of 1:1 (explained above) so, the mass of the sulfate would be:

m K₂SO₄ = 0.36 * 174.26

m K₂SO₄ = 62.73 g

I 'm not sure but the answer might be A.