Question

Number 9.56 questions, a, b , c

Answer 1

Answer:

Explanation:

Chemical equation;

2C₁₀H₂₂ + 31O₂  → 20CO₂ + 22H₂O

a) how many moles of O₂ are needed to completely react with 1.0 mole of C₁₀H₂₂?

Given data:

Moles of C₁₀H₂₂ = 1.0 mol

Moles of O₂ needed = ?

Solution:

we will compare the moles of C₁₀H₂₂ with O₂.

                            C₁₀H₂₂          :            O₂

                                 2              :             31

                                 1.0            :            31/2×1.0 = 15.5 mol

So 15.5 moles of oxygen are needed to react with 1.0 moles of C₁₀H₂₂ .

B) if a car produces 44 g of CO₂, how many grams of C₁₀H₂₂ are used up in the reaction?

Given data:

Mass of CO₂ = 44 g

Mass of C₁₀H₂₂ = ?

Solution:

Number of moles of CO₂:

Number of moles = mass/ molar mass

Number of moles = 44 g/ 44 g/mol

Number of moles = 1 mol

Now we will compare the moles of CO₂ with C₁₀H₂₂.

                          CO₂       :          C₁₀H₂₂

                             20       :              2

                              1         :           2/20×1 = 0.1 mol

Mass of C₁₀H₂₂:

Mass = number of moles × molar mass

Mass = 0.1 mol  × 142.3 g/mol

Mass = 14.23 g

14.23 g of C₁₀H₂₂ are used up in this reaction.

c) if you add 28.8 g of C₁₀H₂₂ to your fuel, how many moles of O₂  are used up in the reaction?

Given data:

Mass of C₁₀H₂₂ = 28.8 g

Moles of oxygen used = ?

Solution:

Number of moles of C₁₀H₂₂ = mass/ molar mass

Number of moles of C₁₀H₂₂ = 28.8 g /142.3 g/mol

Number of moles of C₁₀H₂₂ = 0.20 mol

Now we will compare the moles of C₁₀H₂₂ with oxygen.

                      C₁₀H₂₂            :            O₂

                         2                  :             31

                       0.20               :             31/2×0.20 = 3.1 mol

By adding 28.8 g of C₁₀H₂₂ 3.1 moles of oxygen will used.