The calorimeter above was used to determine the enthalpy of the reaction between

Question

The calorimeter above was used to determine the enthalpy of the reaction between hydrochloric acid and sodium hydroxide. The relevant temperatures at the time of reaction (assuming the reaction was instantaneous) were determined as above. Temperatures were recorded as a function of time and extrapolated to the time of mixing. The generic graph is shown below. The temperature axis is again intentionally not displayed. The relevant information is given in the table below.

The data for the reaction between HCl and NaOH were as follows:

If the temperatures of the acid and base at the time of mixing are different, use their average as the temperature of the mixture before reaction. Use the average also as the initial temperature of the calorimeter.

Question 5

What is the limiting reagent?

A. HCl

B. NaOH

C. The number of moles of acid and base are equal

Enter Your Answer:      A     B     C    

Question 6

How many moles of the limiting reagent reacted?

Enter Your Answer:     

Question 7

What is the average temperature of the mixture of HCl and NaOH at 3.00 min (before they have reacted)?

Enter Your Answer:     

Question 8

What is the weight of the reaction solution in grams?

Enter Your Answer:     

Question 9

What is the change in temperature experienced by the reaction mixture (and the calorimeter) extrapolated back to the time of mixing?

Enter Your Answer:     

Value Units Concentration of HCl 2.26 M Volume of HCl 50.0 mL Density of HCl 1.02 g/mL Concentration of NaOH 2.28 M Volume of NaOH 55.0 mL Density of NaOH 1.02 g/mL Specific Heat of HCl & & NaOH & Reaction Mixture 3.97 J/g-K Temperature of HCl at 3.0 min, Tacid (from graph) 23.9 oC Temperature of NaOH at 3.0 min, Tbase (from graph) 23.9 oC Temperature of reaction mixture at 3.0 min, Tmix  (from graph) 38.9 oC

Explanation / Answer

The reaction is HCl +NaOH-----> NaCl +H2O

Stoichiometric ratio of HCl to NaOH= 1:1

moles of HCl = Molarity* Volume (L)= 2.26*50/1000=0.113 moles

moles of NaOH= Molarity* Volume (L) = 2.28*55/1000 =0.1254

So excess is NaOH and limiting reactant is HCL at 0.113 moles ( A is correct answer)

0.113 moles of limiting reactant HCl reacts

b) Average temperature =23.9 deg.c

c) mass of NaOH= 55*1.02 gm=56.1 gm mass of HCl = 50*1.02= 51 gm

mass of mixture= 56.1+51= 107.1 gm

d) change in temperature = 38.9-23.9= 15 deg.c

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