Question

Information:
temp- 21.5 C
pressure- 1.00 atm
volume- 130 mL of gas

After (.1) Mg was added to HCL bubbled and temp increased to 33.9
pressure- 1.86
volume- 129.9

When letting it cool back to room temp (21.5 C) the new pressure was 1.76 atm

3. Use the ideal gas equation (PV=nRT) to find the number of moles of gas produced by the reaction. The gas constant, R, is given as:

R = 0.082057 (L*atm)/(K*mole)

You will need to convert the temperature from Celsius to Kelvin by adding 273.14
(I'm confused on the ideal gas eq. especially for "n")

Answer 1

n is moles of gas (what you're looking for), and you would use the difference in pressure for P (i.e. 1.76 atm -1 atm).

Answer 2

So we use only the pressures after and before the reaction? and why do you subtract them?

Answer 3

1.76-1 = 0.76 am
n= PV/RT= 0.76 atm x 0.130L /(0.082057 atm L/molK x294.65K)=0.00409 mol
it is not clear in the problem if after the cool down that the volume is equal to the initial volume or
you have to calculate n1 and n2 for the two different conditions before and after the reaction with the formula n= PV/RT then subtract the final minus the initial and you will have the difference in moles of gas produced.
n1=P1V1/RT1 = 1atm x 0.130 L /(0.082057 0.0095961 molesx 294.65K)= 0.0053767 moles
n2=P2V2/RT2 = 1.86 atm x 0.129L/(.0082057atm L/molK x 307.05) = 0.0095961 moles
0.0095961 moles - 0.0053767 moles =0.00415 moles


T P V n R
1 294.65 1 0.13 0.005376768 0.082057
2 307.05 1.86 0.129 0.009523091 0.082057
3 294.65 1.76 0.13082 0.009523091 0.082057

n2-n1= 0.004146323
n=(P2-P1) x V1/(R x T1) 0.004086343

Answer 4

Thank you very much I was very confused as well as which data I was suppose to use. I'm guessing maybe I have to show the other two just as you did. Thanks again you made it very clear for me as this problem was the only one thing in my class so far I've been stuck on. Thanks a ton on your detailed information and the step process. It really helped me allot!