Assuming ideal behavior, how many grams of acetylene are in the tank?

A 75.0 L steel tank at 20.0 deg C contains acetylene gas, C2H2, at a pressure of 1.39 atm.

A) 4.33g B) 6.01g C) 113g D) 1650g

Answer:
T = 273 + 20 = 293 K

pV = nRT
n = pV / RT = 1.39 x 75.0 / 0.0821 x 293 = 4.33 moles of C2H2
Molecular weight = 26 g/mol
4.33 mol x 26 g/mol = 113 g
c)
The ideal gas law is needed

PV = nRT

you are given V (75L); T (20C - remember to convert to Kelvin); P (1.39 atm).

R is a constant (make sure you use the right R for the units of the knowns)

Solve for n - this is the number of moles. It's always about the moles

Multiply by the molecular weight of acetylene to get grams.

Always double check your units - make sure that they cancel so that you are left with grams only.
since you are assuming ideal gas behavior, you can use the eq. PV=nRT, where p=pressure, v=volume, n = #of moles, R=gas constant, and T= temperature.

you just have to plug in the values and solve for n, and convert moles to mass(m.w. = 26). just make sure you choose R in the correct units, 0.0820578437 L · atm · K-1 · mol-1. also, don't forget to convert your temp from C to Kelvin.

when all is said and done, the answer comes out to: c)113g.

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