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3 years ago

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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).?

1 answer:

Anni [7]3 years ago

5 0

Answer:

Explanation:

Given data:

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 data:

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 data:

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 data:

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

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Mass of N2 = 34.7 grams

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