Question

Liquid octane ch3ch26ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 94. g of octane is mixed with 100. g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to 3 significant digits.

Answer 1

Answer:

88.0g of $CO_{2}$

Explanation:

Step 1: Write balanced chemical equation for combustion

$CH_{3}(CH_{2})_{6}CH_{3} + 12.5O_{2} -> 8CO_{2} + 9H_{2}O$

Step 2: Determine moles of octane and oxygen

$n=\frac{m}{M}$

where 'n' is the number of moles, 'm' is the mass and 'M' is the molar mass

$n_{octane}=\frac{m_{octane}}{M_{octane}}$

$n_{octane}=\frac{94}{12*8+1*18}$

$n_{octane}=\frac{94}{114}$

$n_{octane}=0.825 moles$

$n_{oxygen}=\frac{m_{oxygen}}{M_{oxygen}}$

$n_{oxygen}=\frac{100}{2*16}$

$n_{oxygen}=\frac{100}{32}$

$n_{oxygen}=3.125 moles$

Step 3: Determine which is the limiting reactant

1 mole of octane requires 12.5 moles of oxygen for complete combustion

0.825 moles of octane will require 10.3125 moles of oxygen for complete combustion.

As there are not sufficient moles of oxygen for complete combustion of octane, the completion of the reaction will depend on when all moles of oxygen are consumed. Thus, oxygen is the limiting reactant

Step 4: Determine number of moles of carbon dioxide produced

12.5 moles of oxygen will produce 8 moles of carbon dioxide

3.125 moles of oxygen will produce 2.00 moles of carbon dioxide

Step 5: Determine mass of carbon dioxide produced

$m_{carbon dioxide}=n_{carbon dioxide}M_{carbon dioxide}$

$m_{carbon dioxide}=2*(12+2(16))$

$m_{carbon dioxide}=88.0g$

Answer 2

First, we write the balanced reaction equation:

C₈H₁₈ + 13.5O₂ → 8CO₂ + 9H₂O

From this, it is visible that the molar ratio between octane and oxygen should be 1 : 13.5

We check the actual amounts present by first converting them to moles:

Octane:
Mass = 94 g
Molar mass = 114 g/mol
Moles = 94/114 = 0.82

Oxygen:
Mass = 100 g
Molar mass = 32
Moles = 100/32 = 3.12

The molar ratio actually present is 1 : 3.81
This means that oxygen is the limiting reactant and we will base our calculations off of it.

3.12 moles of oxygen will produce:
3.12 * 8/13.5 = 1.85 moles of CO₂

The mass of CO₂ is calculated using:
Moles * molar mass

Mass = 1.85 * 44

The mass of CO₂ produced is 81.4 grams