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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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%