how many moles of hydrogen will be produced from 17.3 g of aluminum in 2Al+3H2SO4=3H2+Al2(SO4)3?

Molar mass of Al = 27.0 g/mol
No. of moles of Al reacted = (17.3 g) / (27.0 g/mol) = 0.6407 mol

2Al + 3H₂SO₄ → 3H₂ + Al₂(SO₄)₃
Mole ratio Al : H₂ = 2 : 3

No. of moles of H₂ produced = (0.6407 mol) × (3/2) = 0.961 mol

2Al + 3H₂SO₄ ➜ 3H₂ + Al₂(SO₄)₃

atomic mass
Al = 27
S = 32
H = 1
O = 16
2Al = 54
3H₂SO₄ = 3(2+32+4•16) = 3•98 = 294
3H₂ = 6
Al₂(SO₄)₃ = 2•27+3•32+12•16 = 342
check 54 + 294 = 6 + 342 = 348

2 mole of Al + 3 moles of H₂SO₄ ➜ 3 mole of H₂ + 1 moles of Al₂(SO₄)₃
54 grams of Al + 294 grams of H₂SO₄ ➜ 6 grams of H₂ + 342 grams of Al₂(SO₄)₃

"How many moles of hydrogen will be produced from 17.3 g of aluminum"

17.3g / 27g/mole = 0.641 mole
ratio is 3 moles H₂ per 2 moles of Al
therefore 0.641 mole Al produces (3/2) 0.641 mole H₂, which is 0.962 mole