Question

100 ml of 0.3 M HCI solution is mixed with 100 ml of 0.33 M KOH solution. The amount of heat liberated is:
Consider,
$△H_{neut}=-55.2kJmo{l}^{-1}$

• A. 1.66 kJ
• B. 1.42 kJ
• C. 1.31 kJ
• D. 1.94 kJ

Answer 1

The correct option is A 1.66 kJ

Amount of HCI added $=\frac{volume\left(inml\right)\times molarity}{1000}$$=\frac{100\times 0.3}{1000}=0.03$ mol

Amount of NaOH added =$=\frac{volume\left(inml\right)\times molarity}{1000}$$=\frac{100\times 0.33}{1000}=0.033$ mol

As 1:1 ratio of both is required for the reaction.
so,
HCl acts as the Limiting reagent
Amount of HCl = 0.03 mol of HCI
So amount of heat evolved $=△H_{neut}\times moles$
$=55.2\times 0.03$
= 1.66 kJ