200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is −57.1 kJ. The increase in temperature in °C of the system on mixing is . The value of x is

Question

200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is −57.1 kJ. The increase in temperature in °C of the system on mixing is . The value of x is (Nearest integer). [Specific heat of water = 4.18 J g⁻¹ K⁻¹; Density of water = 1.00 g cm⁻³] (Assume no volume change on mixing)

Canvas Classes · Accepted Answer

Strategy:
 First, find the limiting reagent and determine the number of moles of water formed. Then, calculate the heat released () and use it to find the temperature change () for the 500 g system.

Step 1: Find moles and limiting reagent
- Moles of .
- Moles of .
Since  is less, it is the limiting reagent. Moles of reaction = 0.03.

Step 2: Calculate heat released ()
- .
- .

Step 3: Calculate 
Total mass = 500 mL = 500 g. Specific heat () = 4.18.

Express as : . Nearest integer: 82.

More questions from Solutions