The complete balance chemical reaction for the compounds stated in the problem is:
C2H6 + (7/2) O2 ---> 2 CO2 + 3 H2O
First, let us find out which is the limiting reactant. We can know by calculating for the moles of reactant.
moles C2H6 = (18 g) (1 mol / 30.07 g) = 0.5986 mol
moles O2 = (85.4 g) (1 mol / 32 g) = 2.6688 mol
The ratio in the chemical reaction is 3.5O2:1C2H6, the given chemicals are in the ratio of:
2.6688O2:0.5986C2H6 = 4.46O2:1C2H6
Therefore this means that there is excess O2 and the C2H6 is our limiting reactant. Therefore we calculate the amount of CO2 based on this.
mass CO2 = (0.5986 mol C2H6) (2 mol CO2/ 1 mol C2H6) (44 g CO2/ mol)
mass CO2 = 52.677g
In 2 significant digits:
mass CO2 = 53 g
The maximum mass of carbon dioxide that could be produced by the reaction will be 53 g
From the equation of the reaction:
2C2H6 +7O2 ---> 4CO2 + 6H2O
Mole of 18 g ethane = 18/30
= 0.6 moles
Mole of 85.4 g O2 = 85.4/32
= 2.67 moles
mole ratio of ethane and oxygen gas = 2:7
Thus, O2 gas seems to be in excess and C2H6 is the limiting reagent.
Mole ratio of C2H6 and CO2 = 1:2
Equivalent mole of CO2 = 0.6 x 2
= 1.2 moles
Mass of 1.2 moles CO2 = 1.2 x 44.01
= 52.81
= 53 g to 2 significant digits
More on stoichiometric calculations can be found here: https://brainly.com/question/8062886
