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.
