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
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?
1 answer
1 answer