Im having a lot of trouble with this question:
H^+ +OH^- --> h2o
Two solutions, 25 mL of 2.5 M HCl and 25 mL of2.5 M NaOH, both initially at 21.1 C are added into a calorimeter and allowed to react. The temperature rose to 37.8 C. Assuming the specific hear of the final solution is 3.8 J/gC and the density of the final solution is 1.04 g/mL
A) calculate the heat of the reaction ignoring qcal.
B)express the unit in heat per mole of the h2o
C) state whether the reaction is endo or exothermic.
Would i use the density to find the mass and then calculate q with this, and then find the limiting reactant to find part B?
H^+ +OH^- --> h2o
Two solutions, 25 mL of 2.5 M HCl and 25 mL of2.5 M NaOH, both initially at 21.1 C are added into a calorimeter and allowed to react. The temperature rose to 37.8 C. Assuming the specific hear of the final solution is 3.8 J/gC and the density of the final solution is 1.04 g/mL
A) calculate the heat of the reaction ignoring qcal.
B)express the unit in heat per mole of the h2o
C) state whether the reaction is endo or exothermic.
Would i use the density to find the mass and then calculate q with this, and then find the limiting reactant to find part B?
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HCl + Na0H = NaCl + H20 Both volumes are the same and both concentrations are the same so both # of moles of NaOH and HCl are the same, so you get complete neutralization.
Temp increase = 37.8 degrees final temp minus 21.1 initial temp = 16.7 degrees
Mass solution 1.04 g/ml times 50 ml = 52 grams.
Heat energy released = 52 grams solution times l6.7 degrees temp increase times 3.8 joules/gm/degree specific heat = 3300 joules released. would be the heat of the reaction.
Moles of water produced. will equal the same as the number of moles of HCl or the number of moles of NaOH since both are equal
Moles of either NaOH or HCl = 2.5 molar times .025 LITERS = .0625 moles of water produced.
heat per mole water. needs a little scaling up.
set this up on a piece of paper
.0625 moles water over 3300 joules released = one mole water over X joules released
Cross multiply and solve for X joules released per mole of water formed.
I got 52800 joules or 52.8 Kj per mole water. Reaction is definitely exothermic.
Temp increase = 37.8 degrees final temp minus 21.1 initial temp = 16.7 degrees
Mass solution 1.04 g/ml times 50 ml = 52 grams.
Heat energy released = 52 grams solution times l6.7 degrees temp increase times 3.8 joules/gm/degree specific heat = 3300 joules released. would be the heat of the reaction.
Moles of water produced. will equal the same as the number of moles of HCl or the number of moles of NaOH since both are equal
Moles of either NaOH or HCl = 2.5 molar times .025 LITERS = .0625 moles of water produced.
heat per mole water. needs a little scaling up.
set this up on a piece of paper
.0625 moles water over 3300 joules released = one mole water over X joules released
Cross multiply and solve for X joules released per mole of water formed.
I got 52800 joules or 52.8 Kj per mole water. Reaction is definitely exothermic.