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

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To find the number of grams of hydrogen in 6.25 L at 1.95 ATM and 243 K, use the ideal gas law equation, pV = nRT. Substituting the known values: (1.95 atm) * (6.25 L) = n * (0.0821 L atm mol-1 K-1) * (243 K). Rearranging gives n = (1.95 * 6.25) / (0.0821 * 243), which calculates to approximately 0.61 moles of hydrogen. To convert moles to grams, multiply by the molar mass of hydrogen (2 g/mol), resulting in about 1.22 grams of hydrogen. This calculation provides the required mass of hydrogen gas under the specified conditions.
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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!
 
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You know PV = nRT, and you are given P, V, R, and T. You even have R in the correct units. Just solve it!
 
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