Answer :
Answer: The mass of product, [tex]H_2O[/tex] is, 36.0 grams.
Explanation : Given,
Mass of [tex]H_2[/tex] = 4.0 g
Mass of [tex]O_2[/tex] = 32.0 g
Molar mass of [tex]H_2[/tex] = 2 g/mol
Molar mass of [tex]O_2[/tex] = 32 g/mol
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{Given mass }H_2}{\text{Molar mass }H_2}[/tex]
[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2.0mol[/tex]
and,
[tex]\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}[/tex]
[tex]\text{Moles of }O_2=\frac{32.0g}{32g/mol}=1.0mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
From the balanced reaction we conclude that
2 mole of [tex]H_2[/tex] react with 1 mole of [tex]O_2[/tex]
From this we conclude that, there is no limiting and excess reagent.
Now we have to calculate the moles of [tex]H_2O[/tex]
From the reaction, we conclude that
2 moles of [tex]H_2[/tex] react to give 2 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]
Molar mass of [tex]H_2O[/tex] = 18 g/mole
[tex]\text{ Mass of }H_2O=(2.0moles)\times (18g/mole)=36.0g[/tex]
Therefore, the mass of product, [tex]H_2O[/tex] is, 36.0 grams.