Balanced equation C2H4 + 3O2 -> 2CO2 + 2H2O 125 grams ethylene (1 mole C2H4/28.052 grams)(2 mole CO2/1 mole C2H4)(44.01 grams/1 moleCO2) = 392 grams CO2 formed ==================
C2h4 + o2 ------------- co2 + h2o c2h4 + 3o2 ------------- 2co2 + 2h2o
yes they are balanced
the balanced equation is Na2S2O3 plus H2O2 yields 2NaOH plus H2S2O3 plus O2.
1 mole gas = 22.4L 1.5mol C2H4 x 22.4L/mol = 33.6L ethane gas (C2H4)
C2H4 + H2O --> C2H5OHReaction balanced at 1:1:1 mole of each compound, so you'll need 0.132 mol C2H4 and this is equal to:0.132 (mol C2H4) * 28 (g/mol C2H4) = 3.696 g C2H4 = 3.70 g C2H4
The only way C2H4 can combust is in the presence of oxygen. All combustion reactions must contain a hydrocarbon and oxygen.
c12h26(i)--->c2h4(g)+______ Yuu shud work iht ouwt :)
C2H4 + 3O2 -> 2CO2 +2H2O 50.0 grams C2H4 (1 mole C2H4/28.052 grams)(2 mole CO2/1 mole C2H4)(44.01 grams/1 mole CO2) = 157 grams CO2 produced
This reaction is an example of a 'hydration' or 'addition' reaction of alkenes: (ethene + water) C2H4 + H2O --> C2H5OH (ethanol) The reaction is catalysed by H+ ions (acid)
Balanced equation C2H4 + 3O2 -> 2CO2 + 2H2O 125 grams ethylene (1 mole C2H4/28.052 grams)(2 mole CO2/1 mole C2H4)(44.01 grams/1 moleCO2) = 392 grams CO2 formed ==================
It is balanced.
C2h4
addition reactions. I was looking up the same question and found it!!!
C12h26 -> c2h4 +
bg;iugi/lo[0'l
C2h4 + o2 ------------- co2 + h2o c2h4 + 3o2 ------------- 2co2 + 2h2o