According to the equation you have 1 mole of C2H4 and 3 moles of O2.
1 • (22.4L / 270L) = 3 • (22.4L / x)
1/270L = 3/x
x = 3(270) / 1
x = 810 L
810 Liters of oxygen will react with 270 liters of ethene (C2H4) at STP
Answer:
can u send the qsn 4rm the start so tt I can help