Question

Liquid octane

CH3CH26CH3

will react with gaseous oxygen

O2

to produce gaseous carbon dioxide

CO2

and gaseous water

H2O

. Suppose 9.1 g of octane is mixed with 48.6 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

molar mass of octane = 114 g/mol
molar mass of O2 = 32 g/mol
number of mol of octane = (given mass)/(molar mass)
= 9.1/114
= 0.080 mol
number of mol of oxygen = 48.6/32
= 1.5 mol

reaction taking place
CH3(CH2)6CH3 + 47/2O2 -- > 8CO2 + 15/2H2O
1 mol of octane required 47/2 mol of O2
0.080 mol of octane required (47/2)*0.080 mol of O2
0.080 mol of octane required 1.9 mol of O2
but we have 1.5 mol of O2
so, O2 is limiting reagent
47/2 mol of O2 react with 1 mol octane
1 mol of O2 react with 2/47 mol octane
1.5 mol of O2 react with (2/47)*1.5 mol octane
1.5 mol of O2 react with 0.06 mol octane
number of octane used = 0.06 mol
number of octane remaining = (0.080-0.06)
= 0.02 mol
mass of octane left = (number of mole of octane left)*(molar mass of octane)
= (0.02*114) g
= 2.28 g
= 2.3 g

Answer : 2.3 g