Question

In a neutralization reaction 30.0mL of 0.234M KOH solution is reacted with 30.0mL of 0.234M HNO3 solution and a temperature rise of 5.6C was observed. Density of base and acid=1.00g/mL and the specific heat of the mixture is 4.18j/g•C
a) write the balanced equation of the reaction
b) calculate the moles of water produced
c) calculate the total heat produced
d) calculate the enthalpy of neutralization of the reaction

Answer 1

Based on the data provided, the equation of the reaction is:

• $KOH + HNO_3 \rightarrow KCl + H_2O$
• moles of water produced = 0.00702 moles
• Heat produced = 1404.48 J
• Enthalpy of neutralization = 200 kJ/mol

What is enthalpy of neutralization?

Enthalpy of neutralization is the heat change when one mole of hydrogen ions from an acid reacts with one mole of hydroxide ions from an alkali to produce one mole of water.

The equation of the reaction of KOH and HNO3 is given below:

$KOH + HNO_3 \rightarrow KCl + H_2O$

1 mole of KOH reacts with 1 mole of HNO3 to produce 1 mole of water.

Volume of KOH = 30 mL = 0.03 L

moles of KOH = 0.03 × 0.234 = 0.00702 moles

Therefore, 0.00702 moles of water are produced

Heat produce = mass × specific heat × temperature difference

Volume of solution = 60 mL

density of solution = 1.0 g/mL

mass of solution = 60 mL × 1.0 g/mL = 60 g

Heat produced = 60 × 4.18 × 5.6

Heat produced = 1404.48 J

Enthalpy of neutralization = 1404.48 J/0.00702 moles

Enthalpy of neutralization = 200068.37 J/mol

Enthalpy of neutralization = 200 kJ/mol

Learn more about enthalpy of neutralization at: brainly.com/question/24939435