Question

100 ml of 0.3 M HCl solution is reacted with 50 ml of 0.6 M
NaOH solution. The heat released in the neutralization
process is
1)57.1 kJ
2)5.7 kJ
3)1.7 kJ
4)2.7 kJ​

Answer 1

Given :

• 100 ml of 0.3 M HCl
• 50 ml of 0.6 M NaOH

To find :

Heat released in neutralization process

Solution :

First of all write balanced chemical equation

HCl + NaOH ➝ NaCl + H₂O

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Now find mole of HCl & NaOH

Moles of NaOH = molarity of NaOH × volume of NaOH {in L}

Moles of NaOH = 0.6 × (50/1000) moles

Moles of NaOH = 30 × 10⁻³ moles

Moles of HCl = molarity of HCl × volume of HCl {in L}

Moles of HCl = 0.3 × (100/1000)

Moles of HCl = 30 × 10⁻³

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Find limiting agent -

According to balanced chemical equation, 1 mole HCl react with 1 mole of NaOH

Therefore HCl and NaOH will react in equal amount.

Also we are given that , Moles of HCl = moles of NaOH = 30 × 10⁻³ moles .

Therefore both will react fully and none is limiting

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Now we know that,

• HCl is strong acid
• NaOH is strong base

Therefore Heat released in neutralization reaction of 1 mole of HCl & 1 mole of NaOH = 57.1 kJ

1 mole ➝ 57.1 kJ

30 × 10⁻³ ➝ (57.1 kJ) × (30× 10⁻³)

➠ 1713 × 10⁻³ kJ

➠ 1.713 kJ

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ANSWER :

Option 3) 1.7 kJ