How many grams of aluminum in required to produce 17.5 g of hydrogen

Mass of Aluminum = 157.5 g

Given

17.5 g of Hydrogen

Required

Mass of Aluminum

Solution

Reaction

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

mol Hydrogen :

= mass : molar mass

= 17.5 g : 2 g/mol

= 8.75

From the equation, mol Al :

= 2/3 x mol H₂

= 2/3 x 8.75

= 5.83

Mass Al :

= 5.83 x 27 g/mol

= 157.5 g

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