Question

According to the following reaction, how many grams of potassium sulfate will be formed upon the complete reaction of 23.8 grams of potassium hydroxide with excess potassium hydrogen sulfate? potassium hydrogen sulfate (aq) + potassium hydroxide (aq) potassium sulfate (aq) + water (l)

Answer 1

Answer: The mass of potassium sulfate that can be produced is 73.88 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For KOH:

Given mass of KOH = 23.8 g

Molar mass of KOH = 56.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of KOH}=\frac{23.8g}{56.1g/mol}=0.424mol$

The chemical equation for the reaction of KOH and potassium hydrogen sulfate follows:

$KHSO_4+KOH\rightarrow K_2SO_4+H_2O$

As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.

KOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of KOH produces 1 mole of potassium sulfate

So, 0.424 moles of KOH will produce = $\frac{1}{1}\times 0.424=0.424moles$ of potassium sulfate

Now, calculating the mass of potassium sulfate from equation 1, we get:

Molar mass of potassium sulfate = 174.26 g/mol

Moles of potassium sulfate = 0.424 moles

Putting values in equation 1, we get:

$0.424mol=\frac{\text{Mass of potassium sulfate}}{174.26g/mol}\\\\\text{Mass of potassium sulfate}=(0.424mol\times 174.26g/mol)=73.88g$

Hence, the mass of potassium sulfate that can be produced is 73.88 grams