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2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2 (g)

How many L of CO2 gas (measured at STP) will be formed during the combustion of 50.0L C2H2 gas
(measured at STP)?

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2 answers

  1. The equation you wrote is not balanced. I've corrected it.
    2 C2H2(g) + 5 O2(g) ==> 4 CO2(g) + 2 H2O (g)
    50 L C2H2 x (4 L CO2/2 L/2 L C2H2) 50 L* 4L/2L = 100 L @ STP
    Note that when gases are involved in the problem and you want volume you can take a shortcut and use L in place of moles. I've done that above. The long way would be
    mols C2H2 = 50/22.4 = 2.232
    mols CO2 = 2.232 mol C2H2 x (4 mols CO2/2 mol C2H2) = 2.232 x 4/2 = 4.464 moles
    Then 4.464 moles x 22.4 L/mol = 100 L @ STP

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  2. thank you.

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