2. Given the reaction of sodium hydroxide with nitric acid: NaOH(aq) HNO3(aq) H2

Question

2. Given the reaction of sodium hydroxide with nitric acid: NaOH(aq) HNO3(aq) H20(l) + NaNO3(aq) ) calculate the theoretical molar heat of neutralization, using the information given in the laboratory manual. a. What is the net ionic equation for the above reaction? b. How many moles of NaNO3 are produced when 50.0 mL of 2.02 M NaOH is added to 50.0 mL of2.OO M HNO3? C. The addition of 50.0 mL of 2.02 M NaOH to 50.0 mL of 2.00 M HNO3 at 22.0 degreeC resulted in a maximum temperature of 36.1 degree C with a total mass of the mixture of 102.3 g. Calculate the experimental molar heat of neutralization. (Remember to use the specific heat of the product when calculating.)

Explanation / Answer

a) The reaction is

NaOH(aq) + HNO3(aq) ===> NaNO3(aq) + H2O

This is an acid base reaction. for the "net" ionic equation, you have to get two reactants that form a different product.

Na(+1)(aq) + OH(-1)(aq) + H(+1)(aq) + NO3(-1)(aq) <-> Na(+1)(aq) + NO3(-1)(aq) +H2O(0)(l)
The net ionic equation is

OH(-1)(aq) + H(+1)(aq) => H2O(l)

b) Write a balanced chemical equation
NaOH(aq) + HNO3(aq) ===> NaNO3(aq) + H2O
50.0 mL ...... 50.0 mL ....... ??? mol
2.02M ......... 2.00M

Clearly, HNO3 is the limiting reactant.

There are 0.100 mol HNO3 and 0.100 mol of NaNO3 will be produced.

c) Neutralization reaction:

H+ + OH- H2O + heat

Q = mcT

T(°C) = 36.1 °C - 22,0 °C = 14.1 °C
m = 102.3 g
c = ?

molar heat of neutralization = Q/moles of product

moles of product:

The limiting reactant is
moles NaOH = (50 mL)*(2.02 M) = 101.0 mmol
moles HHO3 = (50 mL)*(2.00 M) = 100.0 mmol

The HNO3 is the limiting reactant. Therefore, 100.0 mmol of product was produced in the reaction.

molar heat of neutralization = Q/((100 mmol of product)*(1 mol/1000 mmol))

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