Respuesta :
Answer : The mass of [tex]H_2O[/tex] produced will be, 8.1 grams.
Explanation : Given,
Volume of [tex]CH_4[/tex] = [tex]50dm^3=50L[/tex] [tex](1dm^3=1L)[/tex]
Volume of [tex]O_2[/tex] = [tex]10dm^3=10L[/tex]
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of [tex]CH_4[/tex] and [tex]O_2[/tex].
As, 22.4 L volume of [tex]CH_4[/tex] present in 1 mole of [tex]CH_4[/tex]
So, 50 L volume of [tex]CH_4[/tex] present in [tex]\frac{50}{22.4}=2.23[/tex] mole of [tex]CH_4[/tex]
and,
As, 22.4 L volume of [tex]O_2[/tex] present in 1 mole of [tex]O_2[/tex]
So, 10 L volume of [tex]O_2[/tex] present in [tex]\frac{10}{22.4}=0.45[/tex] mole of [tex]O_2[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]O_2[/tex] react with 1 mole of [tex]CH_4[/tex]
So, 0.45 moles of [tex]O_2[/tex] react with [tex]\frac{0.45}{2}=0.225[/tex] moles of [tex]CH_4[/tex]
From this we conclude that, [tex]CH_4[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]H_2O[/tex].
As, 2 moles of [tex]O_2[/tex] react to give 2 moles of [tex]H_2O[/tex]
As, 0.45 moles of [tex]O_2[/tex] react to give 0.45 moles of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]H_2O[/tex].
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(0.45mole)\times (18g/mole)=8.1g[/tex]
Therefore, the mass of [tex]H_2O[/tex] produced will be, 8.1 grams.
