Hello, I need some help with my chemistry lab. In my data I have: I have room T 24.5C
trial 1 Reactant Mg mass 0.2559g moles 0.0105 initial t 23.60c final t 34.10 c
trial 2 Reactant Mg mass 0.2485g moles 0.0102 initial t 22.11 c final t 33.90 c
trial 3 Reactant MgO mass 0.5080g moles 0.1260 initial t 23.996 c final t 27.80 c
trial 4 reactant MgO mass 0.5024g moles 0.1246 initial t 24.068 c final t 27.50 c
1) One of the assumptions is that the heat generated by the reaction is absorbed by pure
water. Calculate the mass of HCl, the mass of Mg in your first trial and the mass of water
for HCl. Determine the percent of water by mass in the system. Does this support the
assumption that the heat is absorbed by water?
Mass of HCl in Trial 1: 0.1L x 3mol/L x 36.458 g/mol = 10.94 g
Mass of Mg in Trial 1: 0.2559 g
Mass of H2O:
This depends on which of the assumptions we go with.
If the density of the HCl solution is taken as 1 g/ml, then:
Mass of solution = 100 ml x 1 g/ml = 100 g of solution
Mass of H2O = Mass of solution – Mass of HCl
= 100 g – 10.94 g = 89.06 g
Percent H2O = 89.06 g / (100 g + 0.2559 g) = 88.83%
On the other hand, if 100 ml of HCl is considered 100 ml of pure water, then:
Mass of H2O = 100 ml (1 g/ml) = 100 g H20
Percent H2O = 100 g / (100 g + 0.2559 g) = 99.74 %
Or, if the known mass of HCl is added in (which kind of conflicts with the
assumption that the solution is pure water), then:
Percent H2O = 100 g / (100g + 0.2559 g + 10.94 g) = 89.93%
The last question asking “Does this support the assumption that the heat is absorbed by
water?” is unclear. What is “this” referring to?
If “this” is referring to the “percent of water by mass in the system”, then the fact that
around 90% of the solution is water would support that assumption.
If, instead, you are supposed to be calculating the heat of the reaction and the capacity of
the water to absorb that heat, then:
H = H[Products] – H[Reactants]
= [-796.88 + 0] – [0 – 2(167.16)] kJ/mol
= -462.56 kJ/mol
For the reaction of 0.0105 moles:
H = (0.0105 mol)(-462.56 kJ/mol)(1000 J/kJ) = -4856.88 J

From the measured temperature change, the heat absorbed by water and the
calorimeter is:


H = (89.06 g)(4.184 J/gºC)(10.5 ºC) + (10 J/ºC)(10.5 ºC)
H = 4017.6 J
This shows that the water present is capable of absorbing the majority of the heat
of the reaction. The remaining 839.3 J (4856.88 J – 4017.6 J) can be attributed to
heat lost to the surrounding environment.
2) The calculation of the enthalpy of reaction assumes that Mg and MgO are the limiting
reagents. Prove that this is indeed the case for both the Mg and MgO experiment.
In the Mg experiment, the reaction was:
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
If Mg is the limiting reagent, then there must be at least twice as many moles of
HCl as there are of Mg. Otherwise, HCl would be the limiting reagent.
Trial 1:
Mg = 0.0105 moles
HCl = (0.1 L)(3 mol/L) = 0.3 moles
Trial 2:
Mg = 0.0102 moles
HCl = (0.1 L)(3 mol/L) = 0.3 moles
In both cases, the number of moles of HCl is more than twice the number of
moles of Mg available for the reaction. This makes Mg the limiting reagent.
In the MgO experiment, the reaction was:
MgO (s) + 2 HCl (aq)  MgCl2 (aq) + H2O (l)
Again, the required ratio of HCl to MgO is 2:1.
Trial 3:
MgO = 0.1260
HCl = 0.3 moles
Trial 4:
MgO = 0.1246 moles
HCl = 0.3 moles
As with the Mg experiment, there is sufficient HCl to use all of the MgO, but
there is not sufficient MgO available to use all of the HCl. This makes MgO the
limiting reagent.