Question

If we start the reaction below with 126.0 g of Al (molar mass of 26.9815 g/mol) and 674.2 g H₂SO₄ (molar mass of 98.07 g/mol), what is the maximum number of grams of H₂ (molar mass of 2.016 g/mol) that can be produced?

2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

How much excess reagent, in grams, will remain at the end of the reaction in the previous part?

Answer 1

Given weight of Al =126.0 g & molar mass of Al = 26.9815 g/mol

Weight of H₂SO₄ = 674.2 g & molar mass of H₂SO₄ =98.07 g/mol

Reaction is

2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

Now calculate num of mole of Al & H₂SO₄

num of mole of Al = 126.0 / 26.9815 =4.6698 mol of Al

num of mole of H₂SO₄ = 674.2 / 98.07 =6.8746 mol of H₂SO₄

Here 4.6698 mol of Al is used up so need of 4.6698 x (3/2) mol of H₂SO₄

i.e.=7.0047 mol but 6.8746 mol H₂SO₄ available so Al is Excess reagent & H₂SO₄ is limiting

one

molar mass of Al₂(SO₄)₃=342.15 g/mol is prepared by using 6.8746 mol H₂SO₄ .

so mol of Al require is = 6.8746 x (2/3) =4.5830 mol of Al

Excess reagent Al remain =4.6698 -4.5830 = 0.0868 mol

Excess reagent Al in g = 0.0868 x molar mass of Al = 0.0868 x 26.9815 = 2.34 g

Excess reagent Al in g = 2.34 g

B) From reaction formation H2 in reaction is of 3 mol so

Maximum number of gram of H2 = 3 x molar mass of H2 = 3 x 2.016 = 6.048 g

Maximum number of gram of H2 = 6.048 g