When 2.50g of hydrochloric acid reacts with 1.50g of potassium hydroxide to yiel
potassium chloride and water, how many grams of potassium chloride are
produced? What is the limiting reactant between hydrochloric acid and potassium
hydroxide ?

1.50  grams of potassium chloride are produced

Potassium hydroxide is the limiting reactant

Explanation:

HCl + KOH→KCl + H₂O

HCl and KOH react on a 1 : 1 mole ratio. They also produce products on a 1 : 1 ratio.

The limiting reagent, in this case, is KOH because it will be fully exhausted in the reaction while the excess HCl will be left remaining.

The remaining HCL will be 2.50 g – 1.50 g

= 1g

Remember because both reactants react at a 1: 1 ratio

In addition, 1.50 g on KCl will be formed in the products

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%