Question

what mass of aluminium hydroxide is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry?

Answer 1

Answer: The mass of aluminium hydroxide needed is 150.774 grams.

Step-by-step explanation:

The chemical equation for the decomposition of aluminium hydroxide follows:

`2Al(OH)_3\rightarrow Al_2O_3+3H_2O`

At STP:

22.4 L of volume is occupied by 1 mole of a gas.

So, 65 L of volume will be occupied by =
`(1)/(22.4L)* 65L=2.9mol`

By Stoichiometry of the reaction:

3 moles of water is produced by 2 moles of aluminium hydroxide

So, 2.9 moles of water is produced by =
`(2)/(3)* 2.9=1.933mol`

To calculate the mass of aluminium hydroxide, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Number of moles of aluminium hydroxide = 1.933 moles

Molar mass of aluminium hydroxide = 78 g/mol

Putting values in above equation, we get:

`1.933mol=\frac{\text{Mass of aluminium hydroxide}}{78g/mol}\\\\\text{Mass of aluminium hydroxide}=150.774 grams`

Hence, the mass of aluminium hydroxide needed is 150.774 grams.