Calculate the volume that 4.7 kg of ethylene gas (C2H4) will occupy at STP.?

At STP, the volume of one mole of a gas is 22.4 liters.
Mass of one mole = 2 * 12 + 4 = 28 grams
n = 4,700 ÷ 28

This is approximately 168 moles.
V = 22.4 * (7,700 ÷ 28) = 3760 liters
I hope this is helpful for you.

Method 1 :

Molar mass of C₂H₄ = (12.0×2 + 1.0×4) = 28.0 g/mol
No. of moles of C₂H₄ = (4.7 × 1000 g) / (28.0 g/mol) = 167.8 mol

Each mole of a gas occupies 22.4 L at STP.
Volume of C₂H₄ = (167.8 mol) × (22.4 L/mol) = 3760 L (to 3 sig. fig.)

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Method 2 :

For the C₂H₄ gas:
Pressure, P = 1.00 atm
No. of moles, n = 167.8 mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = 273 K

Gas law: PV = nRT
Volume of C₂H₄, V = nRT/P = 167.8 × 0.08206 × 273 / 1.00 L = 3760 L (to 3 sig. fig.)