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

Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors. 

                               2 Al(OH)₃    →    Al₂O₃  +  3 H₂O

According to equation at STP,

       67.2 L (3 moles) of H₂O is produced by  =  78 g of Al(OH)₃
So,
                65.0 L of H₂O will be produced by  =  X g of Al(OH)₃

Solving for X,
                                 X  =  (65.0 L × 78 g) ÷ 67.2 L

                                 X  =  75.44 g of Al(OH)₂
Result:
           75.44 g of Al(OH)₂ is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry

Answer 2

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

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 = $\frac{1}{22.4L}\times 65L=2.9mol$

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 = $\frac{2}{3}\times 2.9=1.933mol$

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.