0.8100g MgO(s)(40.3g/mol) was reacted with 52.3 ml 1.00M HCl(density= 1.00g/ml) in the coffee cup calorimeter, and the observed temperature change was +10.4 degree Celsius. write the equation for this reaction, determine the limiting reactant and find difference H for the reaction in kj/mol.
I'm not sure what you mean by "diffference in H" but I'll assume you mean to find delta H rxn in kJ/mol.
MgO + 2HCl ==> H2O + MgCl2
mols MgO = 0.8100/40.3 = about 0.02 but you may want more accuracy than that.(meaning you do it too)
mols HCl = M x L = about 0.052 (again, you do it more accurately).
Convert mols MgO to mols MgCl2. That's 1:1 ratio; therefore, mols MgCl2 = 0.02.
mols HCl converted to mols MgCl2 = 0.052/2 = about 0.026. Therefore, MgO is the limiting reagent.
q = mass H2O x specific heat H2O x dT.
q = 52.3g x 4.184 x 10.4 = about 2276 J/0.02 mol = ? J/mol and convert to kJ/mol
To determine the limiting reactant and find the enthalpy change (ΔH) for the reaction, we need to first write the balanced equation for the reaction involving magnesium oxide (MgO) and hydrochloric acid (HCl):
MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)
Now, let's calculate the moles of MgO and HCl used in the reaction:
Moles of MgO = Mass of MgO / Molar mass of MgO
= 0.8100 g / 40.3 g/mol
≈ 0.02 mol
Moles of HCl = Volume of HCl / 1000 (to convert ml to L) * Molarity of HCl
= 52.3 ml / 1000 * 1.00 mol/L
= 0.0523 L * 1.00 mol/L
= 0.0523 mol
Now, we need to determine the limiting reactant. The balanced equation shows that the stoichiometric ratio between MgO and HCl is 1:2. Therefore, 1 mole of MgO reacts with 2 moles of HCl.
Since we have 0.02 mol of MgO and 0.0523 mol of HCl, we can find the mole ratio between them:
Mole ratio of MgO to HCl = Moles of MgO / Moles of HCl
= 0.02 mol / 0.0523 mol
≈ 0.382
As the mole ratio is less than 1, we can conclude that HCl is the limiting reactant, as there is not enough HCl to completely react with all the MgO.
Next, to find the ΔH (enthalpy change) for the reaction in kJ/mol, we need to use the equation:
ΔH = q / n
Where:
q = heat absorbed or released during the reaction
n = moles of limiting reactant
In this case, the observed temperature change is +10.4 degrees Celsius. To calculate q, we need to use the equation:
q = m * C * ΔT
Where:
m = mass of solution (assume to be the mass of HCl)
C = specific heat capacity of the solution (assumed to be the same as water, which is 4.18 J/g·°C)
ΔT = temperature change in Celsius
Let's calculate the value of q:
q = m * C * ΔT
= (52.3 ml * 1.00 g/ml) * 4.18 J/g·°C * 10.4 °C
= 2173.9848 J
Now, let's calculate ΔH:
ΔH = q / n
= 2173.9848 J / 0.0523 mol
≈ 41548.8 J/mol
To convert this value to kJ/mol, divide by 1000:
ΔH = 41548.8 J/mol / 1000
≈ 41.5 kJ/mol
Therefore, the enthalpy change for the reaction is approximately 41.5 kJ/mol.