Moles H2SO4= .075*.950=.87
moles NaOH= .075*.250=.0187
reaction 2NaOH+H2SO4>>>2H2O+ Na2SO4
so you need two moles NaOH for every mole of sulfuric, so the reaction has excess acid, and you formed .0187 moles water, the product.
heat lost by reaction- massfluid*c*changetemp=.150*4.18*1.70= you do it, check units.
Hr= heat lost /moles product
In a constant‑pressure calorimeter, 75.0mL of 0.950 M H2SO4 was added to 75.0 mL of 0.250 M NaOH. The reaction caused the temperature of the solution to rise from 23.24∘C to 24.94∘C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/(g⋅°C), respectively), what is ΔH for this reaction (per mole of H2O produced)?
3 answers
Did you make a typo with H2SO4 and M of 0.950? I have worked the problem as if this were not a typo. If you meant it to be 0.250 all the following still applies.
H2SO4 + 2NaOH ==> 2H2O + Na2SO424.94 - 23.
mols NaOH = L x M = 0.075 x 0.250= 0.01875
mols H2O produced = 0.01875 = 0.01875
Total volume is 75 mL + 75 mL = 150 mL= 150 grams.
q = heat released = mass H2O x specific heat H2O x (Tfinal-Tinitial) =
150 x 4.184 x (24.94 - 23.24) = I estimate 1000 J but you need to recalculate this and all math in the problem. I've just estimated.
That is approximately 1000 J for 0.0375 mols H2O produced.
dH = about 1000/0.01875 = about 55,000 J or about 55 kJ/mol H2O.
Post your work if you get stuck.
H2SO4 + 2NaOH ==> 2H2O + Na2SO424.94 - 23.
mols NaOH = L x M = 0.075 x 0.250= 0.01875
mols H2O produced = 0.01875 = 0.01875
Total volume is 75 mL + 75 mL = 150 mL= 150 grams.
q = heat released = mass H2O x specific heat H2O x (Tfinal-Tinitial) =
150 x 4.184 x (24.94 - 23.24) = I estimate 1000 J but you need to recalculate this and all math in the problem. I've just estimated.
That is approximately 1000 J for 0.0375 mols H2O produced.
dH = about 1000/0.01875 = about 55,000 J or about 55 kJ/mol H2O.
Post your work if you get stuck.
I didn't mean to overwrite the post by Bob P but the results of each will give you the same answer.
0 answers