Answer:
The correct answer is option a.
Explanation:
[tex]2C_8H_{18}+25 O_2\rightarrow 16CO_2+18H_2O[/tex]
Moles of octane =[tex]\frac{15.0 g}{114.0 g/mol}=0.1315 mol[/tex]
Moles of oxygen gas = [tex]\frac{15 g}{32} g/mol=0.46875 mol[/tex]
According to reaction, 25 moles of oxygen reacts with 2 moles of octane.
Then 0.46875 moles of oxygen will react with:
[tex]\frac{2}{25}\times 0.46875 mol=0.0375 mol[/tex] 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:
[tex]\frac{16}{25}\times 0.46875 mol=0.3 mol[/tex] 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 %
[tex]\% yield=\frac{\text{Experimental yield}}{\text[Theoretical yield}}\times 100[/tex]
[tex]93\%=\frac{x}{13.2 g}\times 100[/tex]
x =12.27 g ≈ 12 g
12 grams of carbon dioxide will be produced in this engine .
