Question #52b92

1 Answer
Mar 16, 2014

The balance chemical equation is
2 C_2C2H_6H6 + 7O_2O2 ---> 4CO_2O2 + 6H_2H2O

as per the equation : 2 moles of C_2C2H_6H6 needs 7 moles of O_2O2 .

moles of C_2C2H_6H6 = Volume of C_2C2H_6H6 / 22.4 L

moles of C_2C2H_6H6 = 16.4 L / 22.4 L = 0.73 mol

as per the molar ratio X mol of C_2C2H_6H6 will need react with 0.98 mol of O_2O2

2 mol of C_2C2H_6H6 /7 mol of O_2O2 =
X mol of C_2C2H_6H6 / 0.98 mol of O_2O2

7.x = 0.98 x 2
7x = 1.96 , x = 1.96/ 7 = 0.28 mol
0.28 moles of C_2C2H_6H6 can react with 0.98 mol of O_2O2.
All of the oxygen will be used to react with 0.28 mol of C_2C2H_6H6 hence it is a limiting reagent. 0.73 - 0.28 = 0.45 moles of C_2C2H_6H6 will remain unused , so it is an excess reagent.

Unused mass of C_2C2H_6H6 =
Unused moles of C_2C2H_6H6 x molar mass of C_2C2H_6H6
= 0.45 mol x 30 g mol^(-1) molโˆ’1 = 13.5 g.

As for the volume of CO_2CO2 produced, we know from the balanced equation that 22 moles of C_2H_6C2H6 will produce 44 moles of CO_2CO2; thus, the moles of CO_2CO2 produced will be

n_(CO_2) = 0.28 * 2 = 0.56nCO2=0.28โ‹…2=0.56 moles

Therefore, from n = V/V_(mol)n=VVmol, we have

V = 22.4 * 0.56 = 12.54 LV=22.4โ‹…0.56=12.54L