A 100.0 mL sample of 0.300 M Ca(OH)2 is mixed with 100.0 mL of 0.700 M H3PO4 in a coffee cup calorimeter.If both solutions were initially at 35.0 °C and the resulting solution had a temperature that was recorded at 39. 7°C, determine the ΔHrxn in units of kJ/mol. Assume the solution has the same specific heat and density as water.
determine the heat released from the solution mass.
heat=(.1+.1)*cwater*(39.7-35)
Now look a the reaction:
3Ca(OH)2+2H3PO4>>Ca3(PO4)2+6H2O
Now looking at the reactants, you started with .03moles Ca(OH)2, and .07 mles H3PO4. So you have excess Ca(OH)2. The phopsporic acid used all .07moles.
So as a result, you got .035 moles of calcium phosphate.
Hr (kJ/mol)=Heatabove/.03
Ooops, that is .035 in the denominator
To determine the ΔHrxn (enthalpy change of the reaction) in kJ/mol, we need to use the equation:
ΔHrxn = q / n
Where:
- ΔHrxn is the enthalpy change of the reaction in kJ/mol
- q is the heat absorbed or released by the reaction in joules (J)
- n is the number of moles of the limiting reactant involved in the reaction
To solve this problem, we will follow these steps:
Step 1: Calculate the heat transferred (q) in joules.
Step 2: Calculate the number of moles of the limiting reactant (n).
Step 3: Convert the heat transferred to kJ/mol by dividing q by n.
Let's move on to each step.
Step 1: Calculate the heat transferred (q) in joules.
To calculate the heat transferred, we will use the equation:
q = m * C * ΔT
Where:
- q is the heat transferred in joules (J)
- m is the mass of the solution in grams (g)
- C is the specific heat capacity of water, which is approximately 4.18 J/(g·°C)
- ΔT is the change in temperature in degrees Celsius (°C)
First, convert the volume of the solution to mass using the density of water (1 g/mL).
Mass of the solution = Volume of the solution * Density of water
= (100.0 mL + 100.0 mL) * 1 g/mL
= 200 g
Next, calculate the change in temperature (ΔT):
ΔT = Final temperature - Initial temperature
= 39.7 °C - 35.0 °C
= 4.7 °C
Now, substitute the values into the equation to calculate q:
q = 200 g * 4.18 J/(g·°C) * 4.7 °C
= 3952.4 J
Step 2: Calculate the number of moles of the limiting reactant (n).
To determine the limiting reactant, we need to compare the number of moles of each reactant involved in the reaction. From the balanced chemical equation:
Ca(OH)2 + 2H3PO4 -> Ca(H2PO4)2 + 2H2O
We can see that 1 mole of Ca(OH)2 reacts with 2 moles of H3PO4.
Number of moles of Ca(OH)2 = 0.300 M * 0.100 L
= 0.030 mol
Number of moles of H3PO4 = 0.700 M * 0.100 L
= 0.070 mol
Since we have 0.030 moles of Ca(OH)2 and 0.070 moles of H3PO4, H3PO4 is the limiting reactant because it has fewer moles.
Therefore, the number of moles involved in the reaction (n) is 0.070 mol.
Step 3: Convert the heat transferred to kJ/mol.
To convert the heat transferred (q) to kJ/mol, divide q by n:
ΔHrxn = q / n
= 3952.4 J / 0.070 mol
≈ 56463.43 J/mol
Lastly, convert the value to kJ/mol by dividing by 1000:
ΔHrxn ≈ 56.46 kJ/mol
Therefore, the ΔHrxn of the reaction is approximately 56.46 kJ/mol.