Question

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​

Answer 1

Answer:
`O_2` is the limiting reagent

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} H_2=(10.0g)/(2g/mol)=5.00moles`

`\text{Moles of} O_2=(20.0g)/(32g/mol)=0.625moles`

The balanced chemical reaction is :

`2H_2+O_2\rightarrow 2H_2O`

According to stoichiometry :

1 moles of
`O_2` require = 2 moles of
`H_2`

Thus 0.625 moles of
`O_2` will require=
`(2)/(1)* 0.625=1.25moles` of
`H_2`

Thus
`O_2` is the limiting reagent as it limits the formation of product and
`H_2` is the excess reagent as it is present more than the required amount.