Methane, a highly flammable gas, is placed in a piston that has a volume of 2.1 L at STP. The piston then compresses the gas to 125 mL. What is the pressure
inside of the piston?

Use the formula:
P1V1/T1N1 = P2V2/T2N2
You can cross out T1,N1,T2, and N2 because you are working with pressure and volume in this equation.
Now, you are left with Boyle’s Law:
P1V1 = P2V2
Substitute the values in
STP = 1 atm for pressure
(1 atm)(2.1L) = P2 (0.125 L) [I converted 125 ml to liters so they would be the same unit]
Now, divide both sides by 0.125 to find P2
P2 = (1 atm)(2.1 L)/(0.125 L)
Liters cancel out
P2 = 16.8 atm

It’s b. mass i searched it up