Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: eleventhxhour on September 30, 2014, 02:33:18 PM
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1) A student placed 50.0mL of 2.05mol/L NaOH in a coffee-cup calorimeter at 20.4°C, and added 50.0mL of 1.20mol/L H2SO4 also at 20.4°C. After quickly stirring the mixture, its temperature rose to 28.2°C. Determine the enthalpy change for the reaction:
2NaOH(aq) + H2SO4(aq) --> 2H2O(l) + Na2SO4(aq)
So,first I calculated q in q=mcdeltaT by using 100g for the two solutions, 4.18J/g°C for c and the change in temperature, which resulted in 3260J
Next I converted the q to deltaH using the equation -q=deltaH. (resulting in -3260J)
After that I found the limiting reactant by first converting each of the solutions into moles (0.1025mol of NaOH and 0.06molH2SO4) and then finding the limiting reactant by using the mole ratios. I found that the limiting reactant was H2SO4. So, I then simply rearranged the equation deltaH = ndeltaHx so that it was isolated for deltaHx, and then subbed in the deltaH (-3260J) and the moles of H2SO4 (0.06mol).
However, I got the wrong answer. I got -54.3kJ while the book says -63.6kJ. What did I do wrong?
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How many moles of NaOH are required to neutralize 0.06 moles of sulfuric acid?
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0.1025mol...?
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Don't guess, just calculate. Forget about the question you already posted, just do the stoichiometry.