Final answer:
a. The moles of CuSO₄ and KOH reacting are 0.0375 moles and 0.255 moles, respectively. b. CuSO₄ is the limiting reactant. c. The ΔH for the reaction is -211.152 kJ/mol.
Step-by-step explanation:
The number of moles of CuSO₄ and KOH can be calculated using their molarity and volume. Since the molarity is given in M (moles per liter), we need to convert the volume from mL to L by dividing by 1000.
a. For CuSO₄:
Moles of CuSO₄ = Molarity of CuSO₄ × Volume of CuSO₄ (in L)
= 0.5 M × (75 mL ÷ 1000) L
= 0.0375 moles of CuSO₄
For KOH:
Moles of KOH = Molarity of KOH × Volume of KOH (in L)
= 3 M × (85 mL ÷ 1000) L
= 0.255 moles of KOH
b. To determine the limiting reactant, we compare the moles of CuSO₄ and KOH involved in the reaction.
Moles ratio of CuSO₄ to KOH = 1:2
Moles of CuSO₄ = 0.0375
Moles of KOH = 0.255
The CuSO₄ is the limiting reactant because it has fewer moles than twice the moles of KOH.
c. To calculate ΔH for the reaction, we can use the formula:
ΔH = q / (moles of limiting reactant)
where q is the heat released or absorbed by the reaction, and is calculated using the equation:
q = mass × specific heat × ΔT
In this case, the specific heat of the solution is the same as the specific heat of water (4.18 J/g°C) and the mass is the sum of the masses of the CuSO₄ solution and the KOH solution. The final temperature (28°C) should be converted to Kelvin by adding 273.15.
ΔT = (28 + 273.15) K - (20.2 + 273.15) K
= 80.95 K
Mass of CuSO₄ = Molarity of CuSO₄ (in mol/L) × Volume of CuSO₄ (in L) × Molar mass of CuSO₄ (159.62 g/mol)
= 0.5 M × (75 mL ÷ 1000) L × 159.62 g/mol
= 5.97 g
Mass of KOH = Molarity of KOH (in mol/L) × Volume of KOH (in L) × Molar mass of KOH (56.11 g/mol)
= 3 M × (85 mL ÷ 1000) L × 56.11 g/mol
= 14.4 g
Assuming the reaction is exothermic, q will be negative.
q = -(4.18 J/g°C) × (5.97 g + 14.4 g) × 80.95 K
= -7908.93 J
ΔH = -7908.93 J / 0.0375 moles of CuSO₄
= -211152 J/mol
ΔH = -211.152 kJ/mol