1 answer

Click to see additional instructions The molar heat of neutralization from hydrochloric acid (HCI) and sodium...

Question:


Click to see additional instructions The molar heat of neutralization from hydrochloric acid (HCI) and sodium hydroxide (NaOH
Click to see additional instructions The molar heat of neutralization from hydrochloric acid (HCI) and sodium hydroxide (NaOH) is -56.1 kJ/mol. Use this accepted value to predict that heat evolved during the neutralization of 168 mL of 3.00 M HCl solution with 168mL of 3.00 M NaOH solution KJ

Answers

Mol of HCl reacting = M(HCl)*V(HCl)
= 3.00 M * 0.168 L
= 0.504 mol
Same is the mol of NaOH reacting

So,
Q = delta H Neutralisation * number of mol
= -56.1 KJ/mol * 0.504 mol
= -28.3 KJ

Answer: -28.3 KJ

.

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