2Na + 2H2O------> 2NaOH + H2 H=-386kJ
If 5g of sodium is added to 200g of water in an open container:
(A) How much heat is absorbed/released int he reaction?
(B)What volume in formed at 22C and 102kPa?
(C) If boiling point elevation constant is 0.52K molal what is the boiling point?
a. Na is the limiting reagent. You should verify that.
mols Na = 5/23 = about 0.22 but that's an estimate.
[386 kJ/(2*23g)] x 5 = kJ heat released.
b. 5 g Na = about 0.22 mols Na and that will produce 0.22/2 = about 0.11 mols H2. Use PV = nRT to solve for V in L. If you want to use 102 kPa as P then R = 8.314
c. About 0.22 mols Na will produce about 0.22 mols NaOH and m = mols/kg solvent = about 0.22/0.200 = ?.
delta T = i*Kb*m
i for NaOH = 2
detla T = 2*0.52*m
To answer these questions, we need to use the molar mass of sodium, the specific heat capacity of water, the ideal gas law equation, and the boiling point elevation constant. Here's how to find the answers:
(A) To calculate the heat absorbed/released in the reaction, we need to use the equation: q = m × c × ΔT, where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
1. Calculate the number of moles of sodium (Na) using its molar mass. The molar mass of sodium is 22.99 g/mol. Therefore, the number of moles of Na is: 5 g / 22.99 g/mol = 0.217 mol.
2. Since the reaction produces 2 moles of NaOH and releases -386 kJ of heat, we need to multiply the moles of Na by the heat of the reaction. The amount of heat released is: 0.217 mol × -386 kJ/mol = -83.962 kJ (Note: The negative sign indicates that heat is released).
Therefore, the heat absorbed/released in the reaction is approximately -83.962 kJ.
(B) To find the volume of H2 gas formed at 22°C and 102 kPa, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin (K).
1. Convert the temperature from Celsius to Kelvin by adding 273.15. So, 22°C + 273.15 = 295.15 K.
2. Calculate the number of moles of H2 using the stoichiometry of the balanced equation. Since 2 moles of H2 are produced for every 2 moles of Na, the number of moles of H2 is also 0.217 mol.
3. Substitute the values into the ideal gas law equation: (102 kPa) × V = (0.217 mol) × (8.31 J/(mol·K)) × (295.15 K).
Solve for V to find the volume of H2 gas formed.
(C) To calculate the boiling point, we need to use the boiling point elevation constant and the molality of the solution. The boiling point elevation equation is: ΔTb = k × m, where ΔTb is the boiling point elevation, k is the boiling point elevation constant, and m is the molality of the solution.
1. Calculate the molality of the solution. Molality (m) is defined as the number of moles of solute (Na) divided by the mass of the solvent (water) in kilograms. The mass of water is 200 g, which is equal to 0.2 kg. So, the molality is: (0.217 mol) / (0.2 kg) = 1.085 mol/kg.
2. Substitute the values into the boiling point elevation equation to calculate the boiling point elevation: ΔTb = (0.52 K · molal) × (1.085 mol/kg).
The boiling point elevation (ΔTb) gives the difference between the boiling point of the solution and the boiling point of the pure solvent (water). To find the boiling point, you need to add the boiling point elevation to the boiling point of pure water (which is 100°C).
I hope this explanation helps you in finding the answers to the given questions!