Question

Problem PageQuestion Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 58.3 g of octane is mixed with 71. g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

We know that ,

Molecular formula of Octane gas = C8H​​18

Molar mass of Octane =114g/mol

Molecular formula of Oxygen = O​​​​​​2

Molar mass of Oxygen =32g/mo

given data

mass of Octane = 58.3g

mass of Oxygen = 71g

Step (1) Calculation for number of mole..

we know that,

​​​​​​number of mole x = mass of x/(molar mass of x)

Number of mole of Octane = 58.3g/(114g/mol)

Number of mole of Octane = 0.5114mol

Number of mole of Oxygen = 71g/(32g/mol)

Number of mole of Oxygen = 2.2187mol

Step(2)

Balance chemical reaction will be given as follows

2C8H18 + 25O​​​​​​2 -------> 16CO2 + 18H2O

From the stoichiometry of balance chemical reaction.

Because

25 mole of Oxygen react with 2 mole of Octane

So,

1mole oxygen react with (2/25) mole of Octane

Hence,

2.2187 mole of Oxygen react with (2/25)2.2187 mole of Octane.

Number of mole of Octane reacted = (2/25)2.2187mol

Number of mole of Octane reacted = 0.1775mol

step(3)

Number of mole of Unreacted Octane = (total number of mole of Octane) - (number of mole of Octane reacted)

Number of mole of Unreacted Octane = (0.5114mol) - (0.1775mol)

Number of mole of Unreacted Octane = 0.3339mol

Step(4) Calculation for mass of Unreacted Octane

we know that,

mass of x = (number of mole of x)(molar mass of x)

mass of Unreacted Octane =(0.3339mol)(114g/mol)

Mass of Unreacted Octane = 38.06g

Answer = 38.06g