Respuesta :
Answer : The mass of [tex]O_2[/tex] needed are, 48 grams.
Explanation :
First we have to calculate the moles of [tex]H_2O[/tex]
[tex]\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}=\frac{54g}{18g/mol}=3mol[/tex]
Now we have to calculate the moles of [tex]O_2[/tex]
The given balanced chemical reaction is:
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]H_2O[/tex] produces from 1 mole of [tex]O_2[/tex]
So, 3 mole of [tex]H_2O[/tex] produces from [tex]\frac{3}{2}=1.5[/tex] mole of [tex]O_2[/tex]
Now we have to calculate the mass of [tex]O_2[/tex]
[tex]\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2[/tex]
Molar mass of [tex]O_2[/tex] = 32 g/mole
[tex]\text{ Mass of }O_2=(1.5moles)\times (32g/mole)=48g[/tex]
Therefore, the mass of [tex]O_2[/tex] needed are, 48 grams.
