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3 years ago

Sulfuric acid reacts with aluminum hydroxide by double replacement.

1 answer:

5 0

a. 
The balanced equation for the reaction between sulfuric acid and aluminium hydroxide is,
 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O

mass of H₂SO₄ = 30.0 g 
Molar mass of H₂SO₄ = 98 g/mol
moles of H₂SO₄ = 30.0 g /98g /mol = 0.306 mol

mass of Al(OH)₃ = 25.0 g 
Molar mass of Al(OH)₃ = 78 g/mol
moles of Al(OH)₃ = 25.0 g/ 78 g/mol = 0.321 mol

Stoichiometric ratio between H₂SO₄ and Al(OH)₃ is 3 : 2

Hence reacted moles of H₂SO₄ = 0.306 mol
reacted moles of Al(OH)₃ = 0.306 mol x (2 / 3) = 0.204 mol

Hence the limiting reactant is H₂SO₄ 

b.
According to the above calculation, the excess reactant is Al(OH)₃. 

The reacted moles of Al(OH)₃ = 0.306 mol x (2 / 3) = 0.204 mol

The added moles of Al(OH)₃ = 0.321 mol

Hence the remaining Al(OH)₃ moles = added moles - reacted moles
= 0.321 mol - 0.204 mol
= 0.117 mol

Molar mass of Al(OH)₃ = 78 g/mol
Remaining mass of Al(OH)₃ = number of moles x molar mass
= 0.117 mol x 78 g/mol
= 9.126 g

c.

The products formed from the reaction between aluminium hydroxide and sulfuric acid are Al₂(SO₄)₃ and H₂O

The limiting reactant is H₂SO₄ 

The stoichiometric ratio between H₂SO₄ and Al₂(SO₄)₃ is 3 : 1
Reacted moles of H₂SO₄ = 0.306 mol
Hence the moles of Al₂(SO₄)₃ formed = 0.306 mol / 3
= 0.102 mol
Molar mass of Al₂(SO₄)₃ = 342 g/mol
Mass of Al₂(SO₄)₃ formed = 0.102 mol x 342 g/mol
 = 34.884 g

The stoichiometric ratio between H₂SO₄ and H₂O is 3 : 6
Reacted moles of H₂SO₄ = 0.306 mol
Hence the moles of H₂O formed = 0.306 mol x (6 / 3)
= 0.612 mol
Molar mass of H₂O = 18 g/mol
Mass of H₂O formed = 0.612 mol x 18 g/mol
 = 11.016 g

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Answer: 31.8 g

Explanation:

To calculate the moles :

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

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$\text{Moles of} C=\frac{30.0g}{12g/mol}=2.5moles$

$Al_2O_3+3C\rightarrow 2Al+3CO$

According to stoichiometry :

1 mole of $Al_2O_3$ require 3 moles of $C$

Thus 0.59 moles of $Al_2O_3$ will require= $\frac{3}{1}\times 0.59=1.77moles$ of $C$

Thus $Al_2O_3$ is the limiting reagent as it limits the formation of product and $C$ is the excess reagent as it is present in more amount than required.

As 1 mole of $Al_2O_3$ give = 2 moles of $Al$

Thus 0.59 moles of $Al_2O_3$ give = $\frac{2}{1}\times 0.59=1.18moles$ of $Al$

Mass of $Al=moles\times {\text {Molar mass}}=1.18moles\times 27g/mol=31.8g$

Thus 31.8 g of $Al$ will be produced from the given masses of both reactants.

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3 years ago

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Explanation:

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