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Chandan Panigrahi posted an Question
October 06, 2020 • 23:52 pm 30 points
  • IIT JAM
  • Chemistry (CY)

Whon aertkain amount of dthylon wal buth 6398 t hoa wa ovoovod.ie hea of comeuhon op thone 19n, th voame of a ca ntd)that onfenodn to th th rraa

whon aertkain amount oF dthylon wal buth 6398 T hoa wa ovoovod.IE hea oF cOmeuHon OP thone 19n, th voame OF a Ca NTD)that Onfenodn to th th rraa

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    Dinesh khalmaniya 1

    Explanation: C2H4 + 3O2 -------> 2 CO2 + 2 H2O C2H4 + 3O2 -------> 2 CO2 + 2 H2OMoles of O2 reacted = (6226/1411) x 3 C2H4 + 3O2 -------> 2 CO2 + 2 H2OMoles of O2 reacted = (6226/1411) x 3Volume of O2 entered at STP = (6226/1411) x 3 x 224 = 296.5 li Hope you understand Keep studying keep learning..!!!

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