The balanced chemical equation is,
2Mg+2HCl→2MgCl+H2↑
<span>C. Carbon. H. Hydrogen. N. Nitrogen. O. Oxygen. P. Phosphorus. <span>S. Sulfur.</span></span>
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 %.
Answer:
0.123 moles of ammonia, can be produced
Explanation:
First of all, we need to determine the reaction:
Ammonia is produced by the reaction of hydrogen and nitrogen.
3H₂(g) + N₂(g) → 2NH₃(g)
Ratio is 3:2. Let's solve the question with a rule of three:
If 3 moles of hydrogen can produce 2 moles of ammonia
Then, 0.37 moles will produce (0.37 . 2) /3 = 0.123 moles