Question

25 ml, 0.1 M Ba(OH)$_2$ react completely with HCl. Find the weight of HCl (in milligram) required?

Hint:

In stoichiometry, always ensure the mole ratio from the balanced equation is used to convert between reactants and products.

Answer 1

Correct Answer: 182.5

Step 1: Write the chemical equation.
Ba(OH)$_2$ (aq) + 2HCl (aq) → BaCl$_2$ (aq) + 2H$_2$O (l)

Step 2: Find the number of moles of Ba(OH)$_2$.
Moles of Ba(OH)$_2$ = 0.1 M × 0.025 L = 2.5 × 10$^{-3}$ moles

Step 3: Calculate moles of HCl required.
From the equation, 1 mole of Ba(OH)$_2$ reacts with 2 moles of HCl. So, moles of HCl = 2 × 2.5 × 10$^{-3}$ = 5 × 10$^{-3}$ moles.

Step 4: Calculate the mass of HCl.
Molar mass of HCl = 36.5 g/mol
Mass of HCl = moles × molar mass = 5 × 10$^{-3}$ × 36.5 = 182.5 milligrams.

Step 5: Conclusion.
The required weight of HCl is 182.5 milligrams.