Answer: [tex]O_2[/tex] is the limiting reagent
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles[/tex]
[tex]\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles[/tex]
The balanced chemical reaction is :
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
According to stoichiometry :
1 moles of [tex]O_2[/tex] require = 2 moles of [tex]H_2[/tex]
Thus 0.625 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.625=1.25moles[/tex] of [tex]H_2[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent as it is present more than the required amount.