In an initial experiment, 50 mL of 1 M HCl are being reacted with50 mL of 1 M Na

Question

In an initial experiment, 50 mL of 1 M HCl are being reacted with50 mL of 1 M NaOH.
1. How many moles of acid are being neutralized?

2. How many moles of water are being produced?

3. Using the hear ot neutralization from the equation, H+ (aq) +OH- (aq)---> H2O(l) + heat, how many joules are beingproduced?

4. What is the total mass (assuming density of water is 1g/mL?

5. If the specifc heat of water is 4.184 J/g*Celsius and using theequation q(solution)= mass of solution * specific heat of solution* delta T, what is the expected temperature rise (delta T)?

Explanation / Answer

The moles of acid are found by 1 mol/L * .05 L = .05 mol ofacid neutralized (M=mol/L), ( 50mL=.05L) this is also the number of moles of water produced. Thisis becuase the reaction is one to one. The heat of neutralization of HCl (aq) by NaOH (aq) is -55.84kJ/mol H2O produced So the number of joules produced is -55.84kJ/mol * .05mol= 2.792 kJ = 2792 J. H2O= 18.01488g/mol NaCl=58.443g/mol.  .05*58.443=2.922 g total mass NaCl the reation.. if you assume that all of the solution is water then itwould be 100 mL * 1g/mL= 100 g of water. However it is not and the exact total mass could be betterfound by a scale. You can estimate by saying that thesalt has about the same density.   Which whendisloved in water it is about 1.01g/mL. So 100 g of solutionit is. 2792=100*4.186 J/g C. *T T=6.7 C.

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