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Click to see additional instructions The molar heat of neutralization from hydrochloric acid (HCI) and sodium hydroxide (NaOH
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Answer #1

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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