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

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 88 g

From the question,

Liquid octane will react with gaseous oxygen to produce carbon dioxide and gaseous water.

To calculate the maximum mass of carbon dioxide that could be produced if 94 g of octane is mixed with 100 g of oxygen,

First, we will write a balanced chemical equation for the reaction

The chemical formula for Octane is CH₃(CH₂)₆CH₃ = C₈H₁₈

The balanced chemical equation for the reaction is

2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

This means 2 moles of octane will react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water

Now, we will determine the number of moles of each reactant present

• For Octane (C₈H₁₈)

Mass = 94 g

Molar mass = 114.23 g/mol

From the formula

$Number\ of\ moles = \frac{Mass}{Molar \ mass}$

Number of moles of octane present = $\frac{94}{114.23}$

Number of moles of octane present = 0.8229 moles

• For Oxygen (O₂)

Mass = 100 g

Molar mass of oxygen = 32 g/mole

∴ Number of moles of oxygen = $\frac{100}{32}$

Number of moles of oxygen = 3.125 moles

From the balanced chemical equation,

2 moles of octane reacts with 25 moles of oxygen

∴ x moles of octane will react with 3.125 moles of oxygen

x = $\frac{3.125 \times 2 }{25}$

x = 0.25 moles

∴ Only 0.25 moles of octane will react with the 3.125 moles of oxygen

Now,

If 2 moles of octane reacts with 25 moles of oxygen to produce 16 moles of carbon dioxide

Then,

0.25 moles of octane will react with 3.125 moles of oxygen to produce (0.25×8) moles of carbon dioxide

0.25 × 8 = 2 moles

∴ 2 moles of carbon dioxide was produced during the chemical reaction

Now, for the mass of carbon dioxide that could be produced

From the formula

Mass = Number of moles × Molar mass

Number of moles of carbon dioxide = 2 moles

Molar mass of carbon dioxide = 44 g/mol

∴ Mass of carbon dioxide = 2 × 44

Mass of carbon dioxide = 88 g

Hence, the maximum mass of carbon dioxide that could be produced by the chemical reaction is 88 g

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