A 45.0-L steel tank at 20.0°C contains acetylene gas,C₂H₂,at a pressure of 1.39 atm.Assuming ideal behavior,how many grams of acetylene are in the tank?

Question

Quizplus · Accepted Answer

Using the ideal gas law, we can solve for the number of moles of acetylene gas in the tank:

PV = nRT

n = PV/RT

where P = 1.39 atm, V = 45.0 L, R = 0.0821 L atm/mol K (universal gas constant), and T = 20°C + 273.15 = 293.15 K.

The molar mass of acetylene (C2H2) is:

2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol

So the mass of acetylene in grams is:

mass = n x molar mass

Substituting the given values and simplifying:

n = (1.39 atm x 45.0 L)/(0.0821 L atm/mol K x 293.15 K) = 1.92 mol

mass = 1.92 mol x 26.04 g/mol = 50.3 g

Therefore, the answer is choice C) 67.8 g.

A 450-L Steel Tank at 20