Respuesta :

dsdrajlin

The pounds of alum produced when 0.26 g of hydrogen was produced are 0.0434 lb.

First, let's convert 0.126 g of hydrogen to moles using its molar mass (2.02 g/mol).

[tex]0.126 g \times \frac{1moo}{2.02 g} = 0.0624 mol[/tex]

Let's consider the steps to make alum (KAl(SO₄)₂⋅12H₂O) from aluminum (Al).

  1. 2 Al(s) + 2 KOH(aq) + 6 H₂O(l) →2 KAl(OH)₄(aq) + 3 H₂(g)
  2. 2 KAl(OH)₄(aq) + H₂SO₄(aq) → 2 Al(OH)₃(s) + K₂SO₄(aq) + 2 H₂O(l)
  3. 2 Al(OH)₃(s) + H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 6 H₂O(l)
  4. K₂SO₄(aq) + Al₂(SO₄)₃(aq) + 24 H₂O(l) → 2 KAl(SO₄)₂⋅12H₂O(s)

To relate H₂ to KAl(SO₄)₂⋅12H₂O, we need to consider the appropriate molar ratios:

  • In step 1, the molar ratio of H₂ to KAl(OH)₄ is 3:2.
  • In step 2, the molar ratio of KAl(OH)₄ to Al(OH)₃ is 2:2.
  • In step 3, the molar ratio of Al(OH)₃ to Al₂(SO₄)₃ is 2:1.
  • In step 4, the molar ratio of Al₂(SO₄)₃ to KAl(SO₄)₂⋅12H₂O is 1:2.

The moles of KAl(SO₄)₂⋅12H₂O produced from 0.0624 moles of H₂ are:

[tex]0.0624 molH_2 \times \frac{2molKAl(OH)_4}{3molH_2} \times \frac{2molAl(OH)_3}{2molKAl(OH)_4} \times \frac{1molAl_2(SO_4)_3} {2molAl(OH)_3} \times \frac{2molAlum}{1molAl_2(SO_4)_3} =0.0416 mol Alum[/tex]

The molar mass of alum is 474.38 g/mol. The mass corresponding to 0.0416 moles is:

[tex]0.0416 mol \times \frac{474.38g}{mol} = 19.7 g[/tex]

Finally, we convert 19.7 grams to pounds using the conversion factor 1 lb = 454 g.

[tex]19.7 g \times \frac{1lb}{454g} = 0.0434 lb[/tex]

The pounds of alum produced when 0.26 g of hydrogen was produced are 0.0434 lb.

You can learn more about molar ratios here: https://brainly.com/question/15973092