Question

When octane (C8H18) is burned in a particular internal combustion engine, the yield of carbon dioxide is 93%. What mass of carbon dioxide will be produced in this engine when 15.0 g of octane (MW = 114.0 g/mol) is burned with 15.0 g of oxygen gas (MW = 32.0 g/mol)? 2C8H18 + 25 O2 --> 16 CO2 + 18 H2O
a.12g b.13g c.21g d.43g e.54g

Answer 1

Answer:

The correct answer is option a.

Explanation:

$2C_8H_{18}+25 O_2\rightarrow 16CO_2+18H_2O$

Moles of octane =$\frac{15.0 g}{114.0 g/mol}=0.1315 mol$

Moles of oxygen gas = $\frac{15 g}{32} g/mol=0.46875 mol$

According to reaction, 25 moles of oxygen reacts with 2 moles of octane.

Then 0.46875 moles of oxygen will react with:

$\frac{2}{25}\times 0.46875 mol=0.0375 mol$ 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:

$\frac{16}{25}\times 0.46875 mol=0.3 mol$ 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 %

$\% yield=\frac{\text{Experimental yield}}{\text[Theoretical yield}}\times 100$

$93\%=\frac{x}{13.2 g}\times 100$

x =12.27 g ≈ 12 g

12 grams of carbon dioxide will be produced in this engine .