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Suppose H2A reacts with ZOH according to this reaction:
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
When 100.0 mL of 0.175 M H2A and 100.0 mL of 0.225 M ZOH, both initially at 20.6 oC, are mixed in a coffee cup calorimeter, the temperature rises to 23.9 oC and solid Z2A forms.
Δ H rxn per mol of Z2A produced.
Oxygen and Hydrogen are the usual elements.
The atomic mass of A is 15.9854 g/mol.
The atomic mass of Z is 35.5680 g/mol
Assume the density of the solution is 1.0 g/mL and the heat capacity is that of water, 4.184 J/goC.
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
When 100.0 mL of 0.175 M H2A and 100.0 mL of 0.225 M ZOH, both initially at 20.6 oC, are mixed in a coffee cup calorimeter, the temperature rises to 23.9 oC and solid Z2A forms.
Δ H rxn per mol of Z2A produced.
Oxygen and Hydrogen are the usual elements.
The atomic mass of A is 15.9854 g/mol.
The atomic mass of Z is 35.5680 g/mol
Assume the density of the solution is 1.0 g/mL and the heat capacity is that of water, 4.184 J/goC.
Answers
Answered by
DrBob222
I suppose you assume that the heat capacity of the solid, Z2A, is ignored.
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
mols H2A initially = M x L = 0.175 M x 0.100 L = 0.0175
mols ZOH initially = 0.225 x 0.100 = 0.0225
The limiting reagent (LR) is ZOH so 0.0225 of ZOH will react with 1/2 x 0.0175 = 0.00875 mols H2A leaving 0.0175 - 0.00875 = ?
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
mols H2A initially = M x L = 0.175 M x 0.100 L = 0.0175
mols ZOH initially = 0.225 x 0.100 = 0.0225
The limiting reagent (LR) is ZOH so 0.0225 of ZOH will react with 1/2 x 0.0175 = 0.00875 mols H2A leaving 0.0175 - 0.00875 = ?
Answered by
DrBob222
oops! I hit the wrong button and posted before I finished the problem.
I suppose you assume that the heat capacity of the solid, Z2A, is ignored.
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
mols H2A initially = M x L = 0.175 M x 0.100 L = 0.0175
mols ZOH initially = 0.225 x 0.100 = 0.0225
The limiting reagent (LR) is ZOH so 0.0225 of ZOH will react with 1/2 x 0.0175 = 0.00875 mols H2A leaving 0.0175 - 0.00875 = ?
q = mass x specific heat x delta T
mass solution = 200 mL H2O with a density of 1.0 g/mL = 200 g and we ignore the ppt of Z2A so
q = 200 g x 4.184 J/g*C x ( 23.9 - 20.6) = about 2761 J but you need to go through these calculations again to get better numbers.
That's 2761 J from 0.0225 mols ZOH to produce 0.0225 mols H2O (or 1/2 that of Z2A).
2761 J for 0.0225 mols H2O = 2761 x 1/0.0225 = 122, 711 J or 122.711 kJ/2 mols H2O = about 61 kJ/mol H2O or per mol Z2A produced. You will get a slightly different value if you recalculate from the beginning.
I suppose you assume that the heat capacity of the solid, Z2A, is ignored.
H2A(aq) + 2ZOH(aq) --> 2H2O(l) + Z2A(s)
mols H2A initially = M x L = 0.175 M x 0.100 L = 0.0175
mols ZOH initially = 0.225 x 0.100 = 0.0225
The limiting reagent (LR) is ZOH so 0.0225 of ZOH will react with 1/2 x 0.0175 = 0.00875 mols H2A leaving 0.0175 - 0.00875 = ?
q = mass x specific heat x delta T
mass solution = 200 mL H2O with a density of 1.0 g/mL = 200 g and we ignore the ppt of Z2A so
q = 200 g x 4.184 J/g*C x ( 23.9 - 20.6) = about 2761 J but you need to go through these calculations again to get better numbers.
That's 2761 J from 0.0225 mols ZOH to produce 0.0225 mols H2O (or 1/2 that of Z2A).
2761 J for 0.0225 mols H2O = 2761 x 1/0.0225 = 122, 711 J or 122.711 kJ/2 mols H2O = about 61 kJ/mol H2O or per mol Z2A produced. You will get a slightly different value if you recalculate from the beginning.
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