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