Respuesta :
Answer : The amount of water will be, 94.86 grams
Solution :
First we have to calculate the moles of [tex]H_2[/tex] and [tex]O_2[/tex].
[tex]\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{10.54g}{2g/mole}=5.27moles[/tex]
[tex]\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{95.10g}{32g/mole}=2.97moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]H_2[/tex] react with 1 mole of [tex]O_2[/tex]
So, 5.27 moles of [tex]H_2[/tex] react with [tex]\frac{5.27}{2}=2.63[/tex] moles of [tex]O_2[/tex]
That means, in the given balanced reaction, [tex]H_2[/tex] is a limiting reagent because it limits the formation of products and [tex]O_2[/tex] is an excess reagent.
The excess reagent remains [tex](O_2)[/tex] = 2.97 - 2.63 = 0.34 moles
Now we have to calculate the moles of [tex]H_2O[/tex].
As, 2 moles of [tex]H_2[/tex] react with 2 moles of [tex]H_2O[/tex]
So, 5.27 moles of [tex]H_2[/tex] react with [tex]\frac{2}{2}\times 5.27=5.27[/tex] 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=(5.27mole)\times (18g/mole)=94.86g[/tex]
Therefore, the amount of water will be, 94.86 grams
