Respuesta :
Answer:
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.
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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