Answer:
24.70g
Explanation:
We'll begin by writing a balanced equation for the reaction between octane and O2 to produce CO2 and H2O. This is illustrated below:
2CH3(CH2)6CH3 + 25O2 —> 16CO2 + 18H2O
Next, let us obtain the masses of octane and O2 that reacted from the balanced equation. This is illustrated below:
Molar Mass of CH3(CH2)6CH3 = 12 + (3x1) + 6[(12 + (2x1)] + 12 + (3x1) = 12 + 3 + 6[14] + 12 + 3 = 12 + 3 + 84 + 12 + 3 = 114g/mol
Mass of CH3(CH2)6CH3 that reacted from the balanced equation = 2 x 114 = 228g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 that reacted from the balanced equation = 25 x 32 = 800g
Next, let us determine which of the reactant is the limiting reactant. This is illustrated below:
From the balanced equation above,
228g of octane required 800g of oxygen for complete Combustion.
Therefore, 8g of octane will require = (8 x 800)/228 = 28.07g of oxygen.
From the calculations made above we can see clearly that oxygen is the excess reactant as the mass of oxygen required (28.07g) is smaller that what was given(i.e 38.9g) from the question and this implies that there are left over mass of oxygen. Therefore, octane is the limiting reactant.
Now we can obtain the theoretical yield of CO2 as illustrated:
The equation for the reaction is given below:
2CH3(CH2)6CH3 + 25O2 —> 16CO2 + 18H2O
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 produced from the balanced equation = 16x44 = 704g
From the balanced equation above,
228g of octane produced 704g of CO2.
Therefore, 8g of octane will produce = (8 x 704)/228 = 24.70g of CO2.
Therefore the theoretical yield of CO2 is 24.70g
