Question

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Liquid octane

CH3CH26CH3

will react with gaseous oxygen

O2

to produce gaseous carbon dioxide

CO2

and gaseous water

H2O

. Suppose 4.57 g of octane is mixed with 8.6 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to

2

significant digits.

final answer in grams 2 sig fig

Answer 1

Molar mass of C8H18,
MM = 8*MM(C) + 18*MM(H)
= 8*12.01 + 18*1.008
= 114.224 g/mol


mass(C8H18)= 4.57 g

use:
number of mol of C8H18,
n = mass of C8H18/molar mass of C8H18
=(4.57 g)/(114.224 g/mol)
= 4.001*10^-2 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 8.6 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(8.6 g)/(32 g/mol)
= 0.2687 mol
Balanced chemical equation is:
2 C8H18 + 25 O2 ---> 18 H2O + 16 CO2


2 mol of C8H18 reacts with 25 mol of O2
for 4.001*10^-2 mol of C8H18, 0.5001 mol of O2 is required
But we have 0.2687 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

According to balanced equation
mol of H2O formed = (18/25)* moles of O2
= (18/25)*0.2687
= 0.1935 mol


use:
mass of H2O = number of mol * molar mass
= 0.1935*18.02
= 3.49 g
Answer: 3.5 g