The theoretical yield of carbon dioxide formed from the reaction of octane and oxygen gas is 24.70g
We'll start by creating a balanced equation to describe how octane reacts with oxygen to create CO₂ and water. Below is an example to help:
25O₂ + 2CH₃(CH₂)6CH₃----------> 16CO₂ + 18H₂O
Let's next take the octane and oxygen reaction masses from the balanced equation. Below is an example to help:
CH₃ Molar Mass (CH₃)
6CH₃ = 12 + (3x1) + 6[(12 + (2x1)] + 12 + (3x1)
= 12 + 3 + 6(14) + 12 + 3
= 12 + 3 + 84 + 12 + 3
= 114g/mol
From the balanced equation, the mass of CH₃(CH₂)6CH₃ that reacted was equal to 2 x 114 grams.
O₂ Molar Mass = 16 x 2 = 32 g/mol
The balanced equation yielded the reaction mass of O₂ as
25 x 32 = 800g.
Let's next identify the limiting reactant among the reactants. Below is an example to help:
From the above-balanced equation, 800g of oxygen was needed for full combustion with 228g of octane.
As a result, 28.07g of oxygen will be needed for 8g of octane, or (8 x 800)/228.
Since the quantity of oxygen required (28.07g) is less than what was provided (i.e. 38.9g) from the inquiry, it is obvious from the calculations done above that oxygen is the surplus reactant. Octane is the limiting reactant as a result.
Now, as shown, we can determine the theoretical CO₂ yield:
The reaction's equation is provided below:
25O₂ + 2CH₃(CH₂)6CH₃ = 16CO₂ + 18H₂O
Molar CO₂ mass equals
12 + (2x16) = 12 + 32 = 44g/mol
The balanced equation yields a mass of CO₂ equal to
16x44 = 704g.
From the above-balanced equation,
704g of CO₂ was created from 228g of octane.
Thus, 8g of octane will result in = (8 x 704)/228 = 24.70g of CO₂ being produced.
Therefore, 24.70g is the theoretical CO₂ yield.
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