Question

Liquid octane CH3(CH2)6CH3 reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. What is the theoretical yield of water formed from the reaction of 10.3g of octane and 55.8g of oxygen gas?

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)= 10.3 g

use:

number of mol of C8H18,

n = mass of C8H18/molar mass of C8H18

=(10.3 g)/(1.142*10^2 g/mol)

= 9.017*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 55.8 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(55.8 g)/(32 g/mol)

= 1.744 mol

Balanced chemical equation is:

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

2 mol of C8H18 reacts with 25 mol of O2

for 9.017*10^-2 mol of C8H18, 1.127 mol of O2 is required

But we have 1.744 mol of O2

so, C8H18 is limiting reagent

we will use C8H18 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 = (18/2)* moles of C8H18

= (18/2)*9.017*10^-2

= 0.8116 mol

use:

mass of H2O = number of mol * molar mass

= 0.8116*18.02

= 14.62 g

Answer: 14.6 g