Answer: 72 g of [tex]H_2O[/tex] is formed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{64}{32}=2moles[/tex]
[tex]2H_2+O_2(g)\rightarrow 2H_2O[/tex]
As [tex]H_2[/tex] is the excess reagent, [tex]O_2[/tex] is the limiting reagent as it limits the formation of product.
According to stoichiometry :
1 mole of [tex]O_2[/tex] produce= 2 moles of [tex]H_2O[/tex]
Thus 2 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 2=4moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=4moles\times 18g/mol=72g[/tex]
Thus 72 g of [tex]H_2O[/tex] is formed.
