Solving for n: How Many Grams of Hydrogen in 6.25L @ 1.95ATM & 243K?

[n108]

If there are 6.25 L of Hydrogen gas at a pressure of 1.95 ATM and a temperature of 243 K, how many grams of hydrogen are there?

I know that the equation for the ideal gas law is:
pV = nRT
p = pressure
V = volume
n = number of moles
R = the gas constant, 0.0821 L atm mol-1 K-1
T = temperature

but I don't know how to fill in n, the number of moles.
this is my unfinished equation:
(1.95 atm) (6.25 L) = (n) (0.00821 atm mol-1 K-1) (243 K)

help, please!

 

[D H]

You know PV = nRT, and you are given P, V, R, and T. You even have R in the correct units. Just solve it!

 

[Blueshift5]

To solve for n, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = (pV) / (RT)
Now, we can plug in the given values for pressure, volume, gas constant, and temperature:
n = (1.95 atm * 6.25 L) / (0.0821 L atm mol-1 K-1 * 243 K)
n = 0.154 mol of hydrogen
To convert moles to grams, we can use the molar mass of hydrogen, which is 2.016 g/mol. Therefore, the total number of grams of hydrogen in 6.25 L at 1.95 ATM and 243 K is:
0.154 mol * 2.016 g/mol = 0.31 grams of hydrogen.
Therefore, there are 0.31 grams of hydrogen in the given conditions.