When a mixture of 10.0 g of acetylene (c2h2 ) and 10.0 g of oxygen, o2 , is ignited, the resultant combustion reaction produces co2 and h2o. how many grams of c2h2 , o2 , co2 and h2o are present after the reaction?

The balanced chemical reaction is expressed as:

C2H2 + 5/2O2 = 2CO2 + H2O

To determine the correct amounts of the substances after the reaction, we need to determine which is the limiting reactant of the given amounts and use this value for the calculations. We do as follows:

10.0 g C2H2 ( 1 mol / 26.04 g ) = 0.38 mol C2H2
10.0 g O2 ( 1 mol / 32 g ) = 0.31 mol O2

0.38 mol C2H2 (5/2 mol O2 / 1 mol C2H2 ) = 0.95 mol O2
0.31 mol O2 ( 1 mol C2H2 / 5/2 mol O2 ) = 0.124 mol C2H2

Therefore, the limiting reactant would be oxygen gas. 

mass of C2H2 left = 10.0 g - 0.124 mol ( 26.04 g/mol) = 6.77 g C2H2
mass of O2 = 0 g
mass of CO2 = 0.31 mol O2 ( 2 mol CO2 / 5/2 mol O2) ( 44.01 g /mol ) = 10.91 g CO2
mass of H2O = 0.31 mol O2 ( 1 mol H2O / 5/2 mol O2) ( 18.02 g /mol ) = 2.23 g H2O