1. calculate the heat evolved when 75.0 ml of a 2.00M hcl reacts with 37.5 ml of

Question

1. calculate the heat evolved when 75.0 ml of a 2.00M hcl reacts with 37.5 ml of a 4.00M naoh if the temperature rises from 20.0°c to 37.9°c (remember you are calculating this by indirect measurement of heat absorbed by solution)
2. how many mole of water are produced by the actual amount of reactants in problem #1
3.what the molar heat of neutralization in the problem #1
hint what should the units be for the value
4. what the net ionic equation for the reaction in problem #1, what the spectator ions
5. based on your answer to problem #3 what should be the heat of the reaction for the below net ionic equation
3oh + 3h >3h20

Explanation / Answer

1) the rise in temperature = 37.9- 20 = 17.9 C

Moles of HCl reacted = Volume x molairty = 0.075 X 2 = 0.15

Moles of NaOH reacted = Volume X molarity = 0.0375 X 4 = 0.15

Volume of solution = 75 + 37.5 = 112.5 mL

Assuming density of solution = 1g/ml

So mass of solution = 112.5 grams

Specific heat of water = 4.18 J / g C

Heat evolved = Mass of water X specific heat X rise in temperature = 112.5 X 4.18 X 17.9 = 8417.48 Joules

2) moles of water produced will be according to equation

NaOH + HCl --> NaCl + H2O

So when one mole of each NaOH and HCl reacts then one mole of H2O is produced

So moles of water produced from reaction of 0.15 moles of each = 0.15 moles

3) The heat of neutralization for 0.15 moles = 8417.48 Joules

So heat from 1 mole of neutraliztion = molar heat of neutralization = 8417.48/ 0.15 = 56116.53 Joules

4) the net ionic equaiton will be

H+ + OH- --> H2O

5) the reaction energy for three moles of reactants = 56116.53 x 3 = 168349.59 Joules = 168.35 KJ

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