When a mixture of 11.0 g of acetylene (C2H2) and 11.0 g of oxygen (O2) is ignited, the resultant combustion reaction produces CO2 and H2O.
a) How many grams of C2H2 are present after the reaction is complete?
b) How many grams of O2 are present after the reaction is complete?
c)How many grams of CO2 are present after the reaction is complete?
d)How many grams of H2O are present after the reaction is complete?
combustion reaction is
2 C2H2 + 5 O2 .....................> 4 CO2 + 2 H2O
In this reaction, 2 mole C2H2 react with 5 mole O2 to form 4 mole CO2 and 2 mole H2O.
11.0 g of acetylene (C2H2) = 11.0 g / 26.0 g / mole = 0.423 mole.
and
11.0 g of oxygen (O2) = 11.0 g / 32.0 g / mole = 0.344 mole.
O2 is the limiting reactant.
a) grams of C2H2 are present after the reaction is complete = (0.423 - 2 / 5 * 0.344) = 0.1375 mole = 0.1375 * 26.0 = 3.575 g
b) grams of O2 are present after the reaction is complete = 0.00 g
c) grams of CO2 are present after the reaction is complete = (4 / 5) * 0.344 = 0.2752 mole = 0.2752 * 44 = 12.1 g
d) grams of H2O are present after the reaction is complete = (2 / 5) * 0.344 = 0.1375 mole = 0.1375 * 18.0 = 2.48 g
When a mixture of 11.0 g of acetylene (C2H2) and 11.0 g of oxygen (O2) is...
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