How many grams of aluminum sulfate: Al2(SO4)3 , would be formed if 250 g of H2SO4 react with aluminum? The reaction is: 2Al + 3H2SO4 > Al2(SO4)3 + 3H2

The mass of Al₂(SO₄)₃ formed if 250 g of H₂SO₄  react with aluminium is 290.82 g.

The chemical equation is represented below:

Al + H₂SO₄ → Al₂(SO₄)₃ + 3H₂

2Al +3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

atomic mass of aluminium = 27 g

molar mass of H₂SO₄ = 32 + 2 + 64 = 98 g

molar mass of Al₂(SO₄)₃  = 54 + 96 + 192 = 342 g

if 3(98 g) of  H₂SO₄ gives 342 g of Al₂(SO₄)₃

250 g of H₂SO₄ will give ?

cross multiply

mass of Al₂(SO₄)₃ = 250 × 342 / 98 × 3

mass of Al₂(SO₄)₃ = 85500 / 294

mass of Al₂(SO₄)₃ = 290.816326531

mass of Al₂(SO₄)₃ ≈ 290. 82 g

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290.8 grams of aluminium sulfate (Al2(SO4)3) will be produced.

Explanation:

Step 1: Data given

Mass of H2SO4 = 250 grams

Molar mass H2SO4 = 98.08 g/mol

Step 2: The balanced equation

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Step 3: Calculate moles H2SO4

Moles H2SO4 = mass H2SO4 / molar mass H2SO4

Moles H2SO4 = 250 grams / 98.08 g/mol

Moles H2SO5 = 2.55 moles

Step 4: Calculate moles Al2(SO4)3

For 2 moles Al we need 3 moles H2SO4 to produce 1 mol Al2(SO4)3 and 3 moles H2

For 2.55 moles H2SO4 we'll have 2.55/3 = 0.85 moles Al2(SO4)3

Step 5: Calculate mass Al2(SO4)3

Mass Al2(SO4)3 = moles Al2(SO4)3 * molar mass

Mass Al2(SO4)3 = 0.85 moles * 342.15 g/mol

Mass  Al2(SO4)3 = 290.8 grams

290.8 grams of aluminium sulfate (Al2(SO4)3) will be produced.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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