2.64 g of solid magnesium metal is placed into 90 mL of a 0.70 mol/L solution of HCl and the following reaction takes place:
Mg (s) + 2 HCl (aq) ----> MgCl₂ (aq) + H₂ (g)
The reaction is carried out in a simple calorimeter. The temperature of the solution changed from 21.0 °C to 25.9 °C.
According to the information, what is the molar enthalpy of reaction, in kJ/mol, with respect to the limiting reagent?.
First we have to determine the limiting reagent. To do that, we have to find the number of moles of each reactant and then we have to compare them to identify the limiting reagent.
We have 2.64 g of Mg. The atomic mass of Mg is 24.31 amu. Then the molar mass of Mg is 24.31 g/mol. Let's find the number of moles of Mg that are in 2.64 g of it.
mass of Mg = 2.64 g
molar mass of Mg = 24.31 g/mol
number of moles of Mg = 2.64 g / (24.31 g/mol)
number of moles of Mg = 0.109 moles
On the other hand, we placed the metal inside 90 mL of a 0.70 M solution of HCl. In that solution we have:
Molarity = moles of HCl / volume of solution in L
Molarity = 0.70 mol/L = 0.70 M
Volume of solution in L = 90 mL * 1 L / 1000 mL = 0.090 L
moles of HCl = 0.70 mol/L * 0.090 L
moles of HCl = 0.063 mol
So, 0.109 moles of Mg are reacting with 0.063 moles of HCl. Let's see which is in excess and which is the limiting reagent.
Mg (s) + 2 HCl (aq) ----> MgCl₂ (aq) + H₂ (g)
According to the reaction, since the coefficient of Mg is 1 and the coefficient of HCl is 2, 1 mol of Mg will react with 2 moles of HCl. We are going to use that relationship to find the number of moles of Mg that will react with 0.063 mol of HCl.
moles of Mg = 0.063 moles of HCl * 1 mol of Mg/(2 moles of HCl)
moles of Mg = 0.0315 moles.
We found that 0.063 moles of HCl only reacts with 0.0315 moles of Mg, but the sample of Mg that we placed inside of the calorimeter had 0.109 moles (or 2.64 g). So it is in excess, and the limiting reagent is HCl.
0.063 moles of HCl ---> limiting reagent
We are asked for the molar enthalpy of reaction. We have to calculate the heat that the solution absorbed, that was released by the reaction.
Qreaction = - Qsolution
Since the reaction occurs in aqueous solution we can consider that it is water.
Csolution = 4.184 J/(g °C)
We also need the mass of the solution. We mixed 2.64 g of Mg and 90 ml of an aqueous solution of HCl. If we consider that the density of the solution is the same as the density of water.
mass of HCl solution = 90 ml * 1 g/ml
mass of HCl solution = 90 g
mass of Mg = 2.64 g
total mass of solution = mass of HCl solution + mass of Mg
total mass of solution = 90 g + 2.64 g
total mass of solution = 92.64 g
And now we can find the heat that is absorbed by the solution:
Qsolution = m * C * ΔT
ΔT = 25.9 °C - 21.0 °C = 4.9 °C
Qsolution = m * C * ΔT
Qsolution = 92.64 g * 4.184 J/(g °C) * 4.9 °C
Qsolution = 1899 J = 1.90 kJ
Finally we can find the heat released by the reaction.
Qreaction = - Qsolution
Qreaction = - 1.90 kJ
And if we refer that value to the limiting reagent (0.063 moles of HCl) to find the molar enthalpy of reaction:
ΔHreaction = -1.90 kJ/(0.063 moles)
ΔHreaction = - 30.16 kJ/mol
Answer: The molar enthalpy of reaction with respect to the limiting reagent is -30.16 kJ/mol.