Question

If 50.0 dm3 of methane, CH4, react with 10.0 dm3 of air, calculate the grams of water produced.

Answer 1

Answer : The mass of $H_2O$ produced will be, 8.1 grams.

Explanation : Given,

Volume of $CH_4$ = $50dm^3=50L$    $(1dm^3=1L)$

Volume of $O_2$ = $10dm^3=10L$

Molar mass of $H_2O$ = 18 g/mole

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

As, 22.4 L volume of $CH_4$ present in 1 mole of $CH_4$

So, 50 L volume of $CH_4$ present in $\frac{50}{22.4}=2.23$ mole of $CH_4$

and,

As, 22.4 L volume of $O_2$ present in 1 mole of $O_2$

So, 10 L volume of $O_2$ present in $\frac{10}{22.4}=0.45$ mole of $O_2$

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

So, 0.45 moles of $O_2$ react with $\frac{0.45}{2}=0.225$ moles of $CH_4$

From this we conclude that, $CH_4$ 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.

Now we have to calculate the moles of $H_2O$.

As, 2 moles of $O_2$ react to give 2 moles of $H_2O$

As, 0.45 moles of $O_2$ react to give 0.45 moles of $H_2O$

Now we have to calculate the mass of $H_2O$.

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(0.45mole)\times (18g/mole)=8.1g$

Therefore, the mass of $H_2O$ produced will be, 8.1 grams.