Question

Calculate the mass of water produced from the reaction of 24.0 g of H2 and 160.0 g of O2. What is the limiting reagent?

Answer 1

Answer:
`O_2` is the limiting reagent, 180 g of water is produced

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} H_2=(24.0g)/(2g/mol)=12moles`

`\text{Moles of} O_2=(160.0g)/(32g/mol)=5moles`

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

According to stoichiometry :

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

Thus 5 moles of
`O_2` will require=
`(2)/(1)* 5=10moles` 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 1 mole of
`O_2` give = 2 moles of
`H_2O`

Thus 5 moles of
`O_2` give =
`(2)/(1)* 5=10moles` of
`H_2O`

Mass of
`H_2O=moles* {\text {Molar mass}}=10moles* 18g/mol=180g`

Thus 180 g of
`H_2O` will be produced from the given masses of both reactants.