now many grams did I calculate? C. How many L of CO2 gas at 1 atm and 22°C would
ID: 1029882 • Letter: N
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now many grams did I calculate? C. How many L of CO2 gas at 1 atm and 22°C would be created? 8. Heat exchange between a solution and a reaction: A 100.0 ml sample of 0.300 M NaOH is mixed with a 100.0 mL sample of M HNO, in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HN03. Assume l ) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.Explanation / Answer
Ans. Step 1: Determine limiting reactant:
Balanced equation: HCl(aq) + NaOH(aq) ----------> NaCl(aq) + H2O
According to the stoichiometry of balanced reaction, 1 mol HCL is neutralized by 1 mol NaOH.
# Moles of HCl taken = Molarity of solution x Volume of solution in liters
= 0.300 M x 0.100 L
= 0.030 mol
Moles of NaOH taken = 0.300 M x 0.100 L = 0.030
Since the moles of NaOH and HCl are equal to each other, they both are completely neutralized by each other. None of them act as limiting reactant.
# Step 2: Total volume of reaction mixture = 100.0 mL + 100.0 mL = 200.0 mL
Taking density of reaction mixture to be 1.00 g/ mL, total mass of reaction mixture = 200.0 g.
Since the calorimetric constant is NOT provided, the heat gained by coffee cup calorimeter is NOT accounted.
# Heat gained by solution during increase in its temperature is given by-
q = m s dT - equation 1
Where,
q = heat change
m = mass of solution
s1 = specific heat of solution
dT = Final temperature – Initial temperature = (T2 – T1)0C
Putting the values in equation 1-
q = 200.0 g x (4.184 J g-10C-1) x (37.0 – 35.0)0C = 1673.6 J
Therefore, total heat released during neutralization of 0.030 mol NaOH = -1673.6 J, the –ve sign indicates that heat is released during neutralization.
# Step 3: Calculate molar enthalpy of neutralization:
Molar enthalpy of neutralization of NaOH = heat released / Moles of NaOH consumed
= -1673.6 J / 0.030 mol
= -55786.67 J/ mol
= -55.786 kJ/ mol
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