1. Say that 150 mL of a 1.50 M NaOH solution with a mass of 151 grams is added t

Question

1. Say that 150 mL of a 1.50 M NaOH solution with a mass of 151 grams is added to 250 mL of a 1.00 M HCl solution with a mass of 252 grams in a coffee cup calorimeter. Before the mixing the temperature of each solution was 22 degrees celcius. After the reaction was complete, the temperature of the mixed solutions was 30.00 degrees celcius. The specific heat of the mixed solutions is essentially the same as water's, 41.184 J/g degrees celcius.

A. What is the balanced chemical equation for this reaction?

B. So, what is the limiting reagent in this reaction and how many moles of it actually reacted?

C. Now, what is the excess reagent in this reactionand how many moles of it actually reacted?

D. What is the value of delta H rxn?

Explanation / Answer

a) The balanced equation is - HCl (aq) + NaOH (aq) ------> NaCl (aq) + H2O (l)

b) Number of mols of HCl given = 252 g/36.5 g/mol = 6.90 mol

Number of mols of NaOH given = 151 g/40 g/mol = 3.775 mol

According to stoichiometry, 6.90 mols HCl required 6.90 mols NaOH for neutralization reaction.

But there is 3.775 mol NaOH, hence we conclude that NaOH is limiting.

Thus - The number of mols of NaOH actually reacted = 3.775 mol

c) Hence the number of mols of HCl reacted = 3.775 mol

Thus- number of mols of HCl left = 6.90 mol - 3.775 mol = 3.125 mol HCl

d) From the given data -

Mass of total solution = (250+150) g *(4.18 J/g) = -1672 J

This is the heat given by 3.775 mols of NaOH.

Hence - delta H(rean) = -1672 J/ 3.775 mol = -442.9 J/mol

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