In lab we mixed HNO3 and NaOH together and measured the heat produced by using a

Question

In lab we mixed HNO3 and NaOH together and measured the heat produced by using a calorimeter.

The data was: 49.89 mL NaOH + 49.82 mL HNO3, delta T (change in temp) 5.438 degrees C. Both concentrations at 1.0M Avergae delta H reaction for the three trials -1.06 x E01 kcal and standard deviation 2.28 E-01 kcal. Specific Heat 0.9601 cal/g degrees C. Density of 1.022 g/mL.

So it is.. HNO3 (aq) + NaOH (aq) --> NaNO3 (aq) + H2O (l)

I need assistance with

6.) Calculate how many moles of acid (HNO3) were present in the acid solution used in reaction. Then how many moles of water were produced in reaction.

7.) Calculate the mass of solution produced in reaction

8.) Use the mass of solution, specific heat of solution, and its change in temperature, to calculate the amount of heat absorbed by the water. q= mass x specific heat x change in temp

9.) Use the equation in the background that relates the heat loss/gain for the solution (qsolution) to the heat loss/gain for the reaction (qreaction), (this is qreaction = -qsolution I think) to calculate the latter.

10.) Use the caluclated value of qreaction and the number of moles of water produced in your reaction to calcuate delta H for this reaction if it produced 1 mole of water, i.e. delta H per mole water. Calculate value to kcal/mol.

Explanation / Answer

I will answer only to paragraph 6), if you want the answers from the other, you attach them to a new question and answer gladly

HNO3...1.0 mol/L *0.04982L=0.04982 mol

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