Calculate the volume that 4.9 kg of ethylene gas (C2H4) will occupy at STP.?

Method 1 : Use gas law

For the C₂H₄ gas :
Mass, m = 4.9 kg = 4900 g
Molar mass, M = (12.0×2 + 1.0×4) g/mol = 28.0 g/mol
Pressure, P = 1 atm
Temperature, T = 273 K
Gas constant, R = 0.08206 L atm / (mol K)

PV = nRT and n = m/M
Then, PV = (m/M)RT
V = mRT/(PM)

Volume,. V = 4900 × 0.08208 × 273 / (1 × 28.0) L = 3920 L


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Method 2 : Use molar volume at STP

Mass of C₂H₄ = 4.9 kg = 4900 g
Molar mass of C₂H₄ = (12.0×2 + 1.0×4) g/mol = 28.0 g/mol
No. of moles of C₂H₄ = (4900 g) / (28.0 g/mol) = 175 mol

Each mole of C₂H₄ occupies 22.4 L of volume.
Volume of C₂H₄ = (175 mol) × (22.4 L/mol) = 3920 L

Volume at STP.....

Simply use the unit-factor method and do it in one calculation. Everyone knows that one mole of an ideal gas at STP occupies a volume of 22.4L.

4900g C2H4 x (1 mol C2H4 / 28.0g C2H4) x (22.4L C2H4 / 1 mol C2H4) = 3900 L C2H4

The answer must be rounded to two significant digits.

E=MC2