A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added
1 answer:
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = =74.8 %.
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Answer:
Here's what I get
Explanation:
1. Sugar
(a) Dissolving in water
The white solid dissolves in water to give a colourless solution. There is no evidence that a new substance is being produced.
(b) Addition of sodium hydroxide
Adding the colourless solution of sodium hydroxide to the colourless sugar solution gives a colourless solution. There is no evidence that a new substance is being produced.
2. Magnesium sulfate
(a) Dissolving in water
The colourless crystals dissolve in water to give a colourless solution. There is no evidence that a new substance is being produced.
(b) Addition of sodium hydroxide
Adding the colourless solution of sodium hydroxide to the colourless solution of magnesium sulfate gives a white precipitate (see image). This is evidence that a new substance is being produced.
