Question

socratic For the following reaction, 25.6 grams of hydrochloric acid are allowed to react with 52.4 grams of barium hydroxide . hydrochloric acid(aq) barium hydroxide(aq) barium chloride(aq) water(l) What is the maximum amount of barium chloride that can be formed

Answer 1

Answer:

63.7g BaCl₂

Explanation:

Based on the chemical reaction:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

2 moles of HCl reacts per mole of barium hydroxide

 

To solve this question, we need to convert the moles of each reactant to moles and, using the chemical reaction, find limiting reactant as follows:

Moles Ba(OH)₂ - Molar mass: 171.34g/mol-

52.4g Ba(OH)₂ * (1mol / 171.34g) = 0.306 moles Ba(OH)₂

Moles HCl - Molar mass: 36.46g/mol-

25.6g HCl * (1mol / 36.46g) = 0.702 moles HCl

For a complete reaction of 0.702 moles HCl are required:

0.702 moles HCl * (1mol Ba(OH)₂ / 2mol HCl) = 0.351 moles of Ba(OH)₂

As there are just 0.306 moles, limiting reactant is Ba(OH)₂

That means the maximum amount of BaCl₂ produced is:

0.306 moles Ba(OH)₂ * (1mol BaCl₂ / 1 mol Ba(OH)₂) = 0.306 moles BaCl₂

In mass -Molar mass BaCl₂: 208.23g/mol-

0.306 moles BaCl₂ * (208.23g / mol) = 63.7g BaCl₂