Question

A sample of 1.6 g of methane (CH4) is completely burnt in 20.00 g of oxygen. The products are carbon dioxide and water. Which is the excess reactant? Which is the limiting reactant? How much of the excess reactant remains unreacted?

Answer 1

Equation of reaction:

CH₄  +  2O₂ →   CO₂   +   2H₂O                 CH₄ = 16g/mole  O₂ = 32g/mole

16         2*32

16g of CH₄  would react completely with 64g Oxygen gas.
Therefore

1.6g of  CH₄  would require 6.4g Oxygen gas          (Divide through by 10)

But from question:

1.6g of  CH₄  was burnt completely in 20g of Oxygen gas.

a) .We can see that the oxygen reactant was in excess as only 6.4g would take part  in the reaction.

b). The limiting reactant is the Methane CH₄, it is the one that is less available.

c.) Excess reactant is Oxygen. Excess by =  20 - 6.4 = 13.6g

Answer 2

Answer :

The excess reactant is, $O_2$

The limiting reactant is, $CH_4$

The mass of excess reactant remains unreacted is, 13.6 grams

Solution : Given,

Mass of $CH_4$ = 1.6 g

Mass of $O_2$ = 20.00 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$.

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{1.6g}{16g/mole}=0.1moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{20.00g}{32g/mole}=0.625moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $CH_4$ react with 2 mole of $O_2$

So, 0.1 moles of $CH_4$ react with $0.1\times 2=0.2$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $CH_4$ is a limiting reagent and it limits the formation of product.

Remaining moles of excess reactant = 0.625 - 0.2 = 0.425 mol

Now we have to calculate the mass of $MgO$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(0.425moles)\times (32g/mole)=13.6g$

Thus, the mass of excess reactant remains unreacted is, 13.6 grams