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Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following
question. If you have 10.0 GRAMS of H2 and 20.0 GRAMS of Oz, which is
the LIMITING REACTANT?
O H2
OO
O H₂O​

Respuesta :

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.

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