Question

Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

Answer 1

Answer : The limiting reactant is $O_2$

Explanation : Given,

Mass of $C_3H_8$ = 30.0 g

Mass of $O_2$ = 75.0 g

Molar mass of $C_3H_8$ = 44 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_3H_8$ and $O_2$.

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{30.0g}{44g/mole}=0.682moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{75.0g}{32g/mole}=2.34moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction we conclude that

As, 5 mole of $O_2$ react with 1 mole of $C_3H_8$

So, 2.34 moles of $O_2$ react with $\frac{2.34}{5}\times 1=0.468$ moles of $C_3H_8$

From this we conclude that, $C_3H_8$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Therefore, the limiting reactant is $O_2$