An unknown amount of helium (He) gas occupies 10.5 L at 1.47 atm pressure and 321 K. What is the mass of helium gas in the container?

Use R = 0.0821 L.atm/(mol.K).

____________ g He. Do NOT enter unit. Report your final answer with 3 SFs.

I AM THE SAME PERSON WHO ASKED YOU THIS QUESTION AND YOU TOLD ME TO SHOW MY WORK AND I DID ITS AT THE BOTTOM. I am just reposting this so you can see it.

I am just entering the numbers in the right column so you will know where I got numbers from.
V = 10.5
P = 1.47
T = 321 K
R = 0.0821
M = 4.003g/mol
Here's the work,, Let me know if I did anything wrong.
N = P1 V1 / RT1 =
(1.47atm)(10.5) = 15.435 / (0.0821) (321) = 26.3541
15.435 / 26.3541 = 0.58567
m = n x m
= (0.58567mol) (4.003g/mol) = 2.34444
Correct answer is 2.34
Am I right @DrBob222?

3 answers

You're absolutely correct. No problem at all. I told you to post your work IF YOU GOT STUCK. Thanks for showing your work.
This looks right to me once I figured out you mean
(1.47atm)(10.5) = 15.435
and
(0.0821) (321) = 26.3541
then
15.435 / 26.3541 = 0.58567
Just a quick note to emphasize what anonymous correctly pointed out but I want to do this in a slightly different way.
Your set up shows numbers that are not equal and you shouldn't do that mathematically. For example, you wrote this.
(1.47atm)(10.5) = 15.435 / (0.0821) (321) = 26.3541
It is true that (1.47 atm)(10.5 L) = 15.435 but that isn't equal to 26.3541 nor is it equal to 15.435 / (0.0821) (321).
I know it is so much easier to write it your way as one makes the calculations but you are making statements that are incorrect mathematically. You shouldn't do that. Being transparent, I did what you're doing when I was a student but I was caught a few times and one of my profs pointed out that I shouldn't do that anymore. In turn, I caught several students before I retired and they don't do that anymore. I don't either. You should make separate lines/statements as anonymous did in his/her response.