Question

Liquid octane (CH3(CH2)6(CH3) will react with gaseous oxygen(O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O) Suppose 2.3 g of octane is mixed with 5.20 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

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)= 2.3 g

use:

number of mol of C8H18,

n = mass of C8H18/molar mass of C8H18

=(2.3 g)/(1.142*10^2 g/mol)

= 2.014*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 5.2 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(5.2 g)/(32 g/mol)

= 0.1625 mol

Balanced chemical equation is:

2 C8H18 + 25 O2 ---> 18 H2O + 16 CO2

2 mol of C8H18 reacts with 25 mol of O2

for 2.014*10^-2 mol of C8H18, 0.2517 mol of O2 is required

But we have 0.1625 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)*0.1625

= 0.117 mol

use:

mass of H2O = number of mol * molar mass

= 0.117*18.02

= 2.108 g

Answer: 2.1 g