Answer:
63.7g BaCl₂
Explanation:
Based on the chemical reaction:
2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
<em>2 moles of HCl reacts per mole of barium hydroxide</em>
<em> </em>
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:
<em>Moles Ba(OH)₂ - Molar mass: 171.34g/mol-</em>
52.4g Ba(OH)₂ * (1mol / 171.34g) = 0.306 moles Ba(OH)₂
<em>Moles HCl - Molar mass: 36.46g/mol-</em>
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
