Question

What mass of H₂O is formed when excess H₂ reacts with 64 g of O₂

2(H2)+(O2)---->2(H2O)
18g
36g
72g
144g

Answer 1

Answer: 72 g of
`H_2O` is formed.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} O_2=(64)/(32)=2moles`

`2H_2+O_2(g)\rightarrow 2H_2O`

As
`H_2` is the excess reagent,
`O_2` is the limiting reagent as it limits the formation of product.

According to stoichiometry :

1 mole of
`O_2` produce= 2 moles of
`H_2O`

Thus 2 moles of
`O_2` will require=
`(2)/(1)* 2=4moles` of
`H_2O`

Mass of
`H_2O=moles* {\text {Molar mass}}=4moles* 18g/mol=72g`

Thus 72 g of
`H_2O` is formed.