Answer : The mass of
produced will be, 8.1 grams.
Explanation : Given,
Volume of
=
![(1dm^3=1L)](https://tex.z-dn.net/?f=%281dm%5E3%3D1L%29)
Volume of
= ![10dm^3=10L](https://tex.z-dn.net/?f=10dm%5E3%3D10L)
Molar mass of
= 18 g/mole
First we have to calculate the moles of
and
.
As, 22.4 L volume of
present in 1 mole of ![CH_4](https://tex.z-dn.net/?f=CH_4)
So, 50 L volume of
present in
mole of ![CH_4](https://tex.z-dn.net/?f=CH_4)
and,
As, 22.4 L volume of
present in 1 mole of ![O_2](https://tex.z-dn.net/?f=O_2)
So, 10 L volume of
present in
mole of ![O_2](https://tex.z-dn.net/?f=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](https://tex.z-dn.net/?f=CH_4%2B2O_2%5Crightarrow%20CO_2%2B2H_2O)
From the balanced reaction we conclude that
As, 2 moles of
react with 1 mole of ![CH_4](https://tex.z-dn.net/?f=CH_4)
So, 0.45 moles of
react with
moles of ![CH_4](https://tex.z-dn.net/?f=CH_4)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
.
As, 2 moles of
react to give 2 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
As, 0.45 moles of
react to give 0.45 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Now we have to calculate the mass of
.
![\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20%7DH_2O%3D%5Ctext%7BMoles%20of%20%7DH_2O%5Ctimes%20%5Ctext%7BMolar%20mass%20of%20%7DH_2O)
![\text{Mass of }H_2O=(0.45mole)\times (18g/mole)=8.1g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20%7DH_2O%3D%280.45mole%29%5Ctimes%20%2818g%2Fmole%29%3D8.1g)
Therefore, the mass of
produced will be, 8.1 grams.