A student in chemistry 150-02 weighed out 55.5 g of octane (C8H18) and allowed it to react with oxygen, O2. The products formed were carbon dioxide (CO2) and water (H2O).

Write a balanced equation for the reaction

How many grams of oxygen are required to react with 55.0g of octane (C8H18) ?

How many grams of CO2 are produced from 55.0g of octane (C8H18) ?

How many molecules of H2O are produced from 55.0g of octane (C8H18) ?

How many gams of C8H18 are required to produce 30.0g of water (H2O).?

Given dа ta:

Mass of octane = 55.5 g

Balanced chemical equation = ?

Mass of oxygen required to react = ?

Mass of CO₂ for med = ?

Molecules of water produced = ?

Mass of octane required to produced 30.0 g of water = ?

Solution:

1)

Chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

2)

Mass of oxygen required to react = ?

Mass of octane = 55.0 g

Solution:

Number of moles of octane:

Number of moles = mass / molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with oxygen.

C₈H₁₈ : O₂

2 : 25

0.48 : 25/2*0.48 = 6 mol

Mass of oxygen required:

Mass = number of moles * molar mass

Mass = 6 mol * 32 g/mol

Mass = 192 g

3)

Given dа ta:

Mass of carbon dioxide produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass / molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with CO₂.

C₈H₁₈ : CO₂

2 : 16

0.48 : 16/2*0.48 = 3.84 mol

Mass of CO₂ produced:

Mass = number of moles * molar mass

Mass = 3.84 mol * 44 g/mol

Mass = 168.96 g

4)

Given dа ta:

Molecules of water produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass / molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with H₂O.

C₈H₁₈ : H₂O

2 : 18

0.48 : 18/2*0.48 = 4.32 mol

Number of molecules of water:

1 mol = 6.022 * 10²³ molecules

4.32 mol * 6.022 * 10²³ molecules / 1 mol

26 * 10²³ molecules

5)

Given dа ta:

Mass of octane required = ?

Mass of water produced = 30 g

Solution:

Number of moles of water.

Number of moles = mass / molar mass

Number of moles = 30 g / 18 gmol

Number of moles = 1.67 mol

Now we will compare the moles of water and octane from balance chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

H₂O : C₈H₁₈

18 : 2

1.67 : 2/18*1.67 = 0.185 mol

Mass of octane:

Mass = number of moles * molar mass

Mass = 0.185 * 114.23 g/mol

Mass = 21.13 g