Response:
9.9 ml of 0.200M NH₄OH(aq)
Reasoning:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?
1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution
1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)
=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters
Answer:
1219.5 kJ/mol
Explanation:
The calculation for this value requires using the following equation:
ΔHºrxn = Σn * (BE of reactants) - Σn * (BE of products)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H)].
The bond energy (BE) values are:
BE C = C: 839 kJ/mol
BE C-H: 413 kJ/mol
BE O = O: 495 kJ/mol
BE C = O: 799 kJ/mol
BE O-H: 463 kJ/mol
By substituting these values into the equation, you will get:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463)] = 1219.5 kJ/mol
The molality of sodium chloride comes to 2.55 mol/kg.
V(solution) = 100 ml.
The mass of the solution (m) is calculated using the formula: m(solution) = d(solution) · V(solution).
Here, m(solution) would be 1.10 g/ml multiplied by 100 ml.
Thus, m(solution) equals 110 g.
The mass fraction of NaCl (ω(NaCl)) is 13.0% or 0.13.
Substituting this into the formula gives m(NaCl) = ω(NaCl) · m(solution).
This results in m(NaCl) equaling 0.13 multiplied by 110 g.
Concluding with m(NaCl) equals 14.3 g.
The number of moles of NaCl (n(NaCl)) is calculated by n(NaCl) = m(NaCl) ÷ M(NaCl).
So, n(NaCl) equals 14.3 g divided by 58.5 g/mol.
This results in n(NaCl) = 0.244 mol.
The mass of water (m(H₂O)) is found as 110 g minus 14.3 g.
Thus, m(H₂O) equals 95.7 g which converts to 0.0957 kg.
Finally, b(NaCl) is calculated as n(NaCl) ÷ m(H₂O).
Hence, b(NaCl) = 0.244 mol divided by 0.0957 kg.
This yields b(NaCl) = 2.55 mol/kg.
