Final answer:
The maximum amount of potassium sulfate that can be formed is 44.1 grams. KH(SO4) is the limiting reagent, and after the reaction, 5.9 grams of the excess reagent KOH remains.
Step-by-step explanation:
The student is asking about a chemical reaction in which potassium hydrogen sulfate reacts with potassium hydroxide to form potassium sulfate and water. To determine the maximum amount of potassium sulfate that can be formed, we need to conduct stoichiometry calculations based on the molecular weights of the reagents and the balanced chemical equation:
KH(SO4) + KOH → K2SO4 + H2O
First, we calculate the moles of each reactant:
- 42.1 g KH(SO4) / (39.1+1+32.1+(16x4)) g/mol = 0.258 moles of KH(SO4)
- 20.6 g KOH / (39.1+16+1) g/mol = 0.349 moles of KOH
According to the balanced equation, they react in a 1:1 mole ratio. Therefore, KH(SO4) is the limiting reagent because it has fewer moles. Using the limiting reagent's moles, we find the maximum amount of K2SO4 formed:
0.258 moles K2SO4 x (39.1x2+32.1+(16x4)) g/mol = 44.1 grams of K2SO4
The excess reagent KOH would have:
0.349 moles KOH - 0.258 moles KH(SO4) = 0.091 moles KOH leftover
0.091 moles KOH x (39.1+16+1) g/mol = 5.9 grams of KOH remaining