c you have right but for the wrong reason. You don't IGNORE the water vapor, you count it.
Ptotal = pH2 + pH2O
pH2 = Ptotal - pH2O
For part e, you don't distill anything. You dilute 6.0M HCl so its final concn is 1.0M.
mL1 x M1 = mL2 x M2
50 mL x 1.0M = mL2 x 6.0
mL2 = 50 x 1/6 = about 8.33
So you use the graduated cylinder to measure out 8.3 mL of the 6.0M HCl, pour that into the 50 mL volumetric flask, add distilled H2O to the mark in the flask, and mix thoroughly.
I don't know if this is accurate enough for your prof; I don't know how accurate you are to be. If you need more accuracy than you can get as described above, you can calibrate the dropper and add the appropriate number of drops to obtain 8.33 mL (or at least as close as that will permit).
Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g)
An experiment is performed to determine the volume of hydrogen gas produced when calcium reacts with excess HCl(aq) as represented by the net ionic equation above. A 0.0720 g sample of pure calcium and a solution of 1.0 M HCl(aq) were used in the experiment.
(a) Calculate the number of moles of calcium in the 0.0720 g sample.
My answer: 0.001796 moles Ca
(b) Calculate the number of moles of HCl(aq) needed to react completely with the sample of calcium.
My answer: 0.003592 moles HCl
As the calcium reacts, the hydrogen gas produced is collected by water displacement at 25°C. The pressure of the gas in the collection tube is measured to be 752 torr.
(c) Given that the equilibrium vapor pressure of water is 24 torr at 25.0°C, calculate the pressure that the H2(g)
produced in the reaction would have if it were dry.
(d) Calculate the dry volume, in liters measured at the conditions in the laboratory, that the H2(g) would have
produced in the reaction.
(e) The laboratory procedure specified 1.0 M HCl solution be, but only 6.0 M HCl solution was available. Describe
the steps for safely preparing 50.0 mL of 1.0 M HCl(aq) using 6.0 M HCl solution and materials selected from
the list below. Show any necessary calculation(s).
10.0 mL graduated cylinder
Distilled water
250 mL beakers
Balance
50.00 mL volumetric flask
Dropper
For parts (c) and (d), when it says "dry," does that mean ignore the presence of the water vapor? So for (c), calculate the pressure of the gas without the equilibrium vapor pressure of water? Would you just subtract 24 torr from 752 torr? For (d), how does you find the "dry volume"?
For part (e), how much water would you add to distill the 6.0M HCl to 1.0M HCl?
1 answer