A student performed Part A of the experiment and obtained the following data: Ex

Question

A student performed Part A of the experiment and obtained the following data: Exothermic Reaction a) Initial temperature in calorimeter _____________ degree C b) Final temperature in calorimeter _____________ degree C c) Volume of 1.0 M NaOH used _______________ mL d) Volume of 1.0 M HCl used _______________ mL What is the total mass in the calorimeter? (Keep in mind that the density of the solutions are both 1.000 g/mL, so this makes the number of mL equal to the number of grams). __________g What is delta T for this data? (delta T = T_final -T_initial) __________degree C Calculate q. __________Joule From the balanced equation, we see that the numbers of moles of H_2 O is a 1 to 1 mole ratio to the number of moles of acid and base. What is the number of moles in 49.8 mL of 1.0 M NaOH ? (Use M = mol/L) ___________mol What is the number of moles of H_2 O formed? __________mol Why was the moles of NaOH used to find the moles of H_2 O formed, and not the moles of HCI? Calculate the heat per mole, delta H_rxn _________J/mol

Explanation / Answer

1) Total mass in calorimeter = 50.1+ 49.8 = 99.9 g

2) delta T = 31.8-23.3 = 8.5

3) q = heat absorbed by calorimeter = mass x specific heat x rise in temperature

= 99.9g x 4.18J/g x 8.5

= 3549.447 J

4) NaOH + HCl -------> NaCl + H2O

Moles of NaoH used = 49.8x 1 /1000 = 0.0498 moles

and moles of HCl used = 50.1 x1/1000 = 0.0501 moles

5) the number of moles of H2O formed = moles of NaOH = 0.0498 moles

60 The number of moles of NaOh are less than moles of HCl. Thus it is the limiting reagent and some HCl is remained unreacted.

7) If 0.0498 moles of base on neutralization can produce 3549.447 J

1 mole of acid-base rxn can produce = 3549.447Jx1 mol / 0.0498mol

= 71274.0J /mol

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