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

To solve the problem of finding the weight of HCl required to react completely with 25 ml of 0.1 M Ba(OH)₂, we follow these steps:

Step 1: Write the balanced chemical equation.
The reaction between Ba(OH)₂ and HCl is:
Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
This equation shows that 1 mole of Ba(OH)₂ reacts with 2 moles of HCl.

Step 2: Calculate the number of moles of Ba(OH)₂.
The concentration of Ba(OH)₂ is 0.1 M, and the volume is 25 ml (0.025 L).
The number of moles of Ba(OH)₂ is given by:
n = C × V = 0.1 mol/L × 0.025 L = 0.0025 mol

Step 3: Determine the number of moles of HCl required.
According to the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HCl. Therefore,
n(HCl) = 2 × n(Ba(OH)₂) = 2 × 0.0025 mol = 0.005 mol

Step 4: Calculate the weight of HCl in milligrams.
The molar mass of HCl is approximately 36.46 g/mol. Therefore,
Weight of HCl = n(HCl) × Molar Mass of HCl
= 0.005 mol × 36.46 g/mol = 0.1823 g
Convert 0.1823 g to milligrams:
0.1823 g × 1000 mg/g = 182.3 mg

Step 5: Validate the solution.
The calculated weight of HCl is 182.3 mg, which falls within the specified range of 182.5 mg (given as a specific expected value). While the calculated 182.3 mg slightly deviates from 182.5 mg, it is a reasonable result considering possible rounding in intermediate steps or provided data precision.

Thus, the weight of HCl required is approximately 182.3 mg.