absolutely incorrect.
1. Subtract the water vapor pressure from the observed pressure to get the pressure of the hydrogen gas.
2. figure the moles of that hydrogen at its pressure, temp, and volume.
3. According to the balanced equation, that should also be the same number of moles of sodium hydride. Convert that to grams of NaH
Here is the question:
sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas:
NaH(s)+H2O-->NaOH(aq)+H2(g)
A sample of NaH weighing ________g will produce 982mL of gas at 28.0 C and 765 torr, when the hydrogen is collected over water. The vapor pressure of water at this temperature is 28 torr.
Note: You may have to consider that all the pressure stated above does not solely belong to the hydrogen gas.
Here is my Homework:
n=PV/RT
P=1.01atm or 765torr
R=.0821
V=.982 L
T=301.15 K
(1.01atm H2)(.989L H2)/(.0821 Latm/mol L)(301.15k H2)==3638.08 mol H2
3638.08 mol H2 = 3638.08 mol NaH
(3638.08 mol NaH)23.99771 g/mol NaH=87305.5887968
The assessment is saying this answer is incorrect can anyone tell me what I am doing wrong?
4 answers
Ok so I took the 1.01 atm H2 and subtracted the .03684atm H2O=.97316
Then put that in the equation I did earlier.
n=PV/RT
P=.97316
R=.0821
V=.982 L
T=301.15 K
(.97316atm H2)(.989L H2)/(.0821 Latm/mol L)(301.15k H2)==3505.383 mol H2
3505.383 mol H2 = 3505.383 mol NaH
(3505.383 mol NaH)23.99771 g/mol NaH=84121.373
and the assesment is still saying this is wrong??? some times the assesments answers have to be with in a very limite tolerance but I carried the sig figs through quite well so I think I still made another mistake?
Then put that in the equation I did earlier.
n=PV/RT
P=.97316
R=.0821
V=.982 L
T=301.15 K
(.97316atm H2)(.989L H2)/(.0821 Latm/mol L)(301.15k H2)==3505.383 mol H2
3505.383 mol H2 = 3505.383 mol NaH
(3505.383 mol NaH)23.99771 g/mol NaH=84121.373
and the assesment is still saying this is wrong??? some times the assesments answers have to be with in a very limite tolerance but I carried the sig figs through quite well so I think I still made another mistake?
Yes, think on this: At STP one mole of any gas will occupy 22.4L. Your conditions are near STP, so 1 Liter ought to be about .05moles, not three thousand.
Examining your calcs, the numerator is about one, the denominator is about 24 (in my head), that is about .04 moles. Redo your calcs.
Examining your calcs, the numerator is about one, the denominator is about 24 (in my head), that is about .04 moles. Redo your calcs.
Thank you so much for sticking with me on this, for some reason I am putting the entire problem into my TI-83 and it is coming up with those large numbers. I do not get why it is doing that but now I got the right answer.