Answer:
The correct answer is option a.
Explanation:
Moles of octane =
Moles of oxygen gas =
According to reaction, 25 moles of oxygen reacts with 2 moles of octane.
Then 0.46875 moles of oxygen will react with:
of octane
Oxygen is present in limited amount. Hence, limiting reagent.
According to reaction , 25 moles of oxygen gives 16 moles of carbon-dioxide.
Then 0.46875 moles of oxygen will give:
of carbon-dioxide
Mass of 0.3 moles of carbon-dioxide:
0.3 g × 44 g/mol = 13.2 g
Theoretical yield = 13.2 g
Experimental yield = x
Percentage yield of carbon dioxide = 93 %
x =12.27 g ≈ 12 g
12 grams of carbon dioxide will be produced in this engine .