Question

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

CH4 (g) + 2 O2 (g) --> CO2 (g) + 2 H2O (l)

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
`(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
`(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
`(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* \text{Molar mass of }H_2O`

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

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