Respuesta :
Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors.
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
2 Al(OH)₃ → Al₂O₃ + 3 H₂O
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: The mass of aluminium hydroxide needed is 150.774 grams.
Explanation:
The chemical equation for the decomposition of aluminium hydroxide follows:
[tex]2Al(OH)_3\rightarrow Al_2O_3+3H_2O[/tex]
At STP:
22.4 L of volume is occupied by 1 mole of a gas.
So, 65 L of volume will be occupied by = [tex]\frac{1}{22.4L}\times 65L=2.9mol[/tex]
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 = [tex]\frac{2}{3}\times 2.9=1.933mol[/tex]
To calculate the mass of aluminium hydroxide, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Number of moles of aluminium hydroxide = 1.933 moles
Molar mass of aluminium hydroxide = 78 g/mol
Putting values in above equation, we get:
[tex]1.933mol=\frac{\text{Mass of aluminium hydroxide}}{78g/mol}\\\\\text{Mass of aluminium hydroxide}=150.774 grams[/tex]
Hence, the mass of aluminium hydroxide needed is 150.774 grams.
