Question
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?
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?
Answers
DrBob222
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
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