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Aluminum reacts with sulfuric acid to produce aluminum sulfate in hydrogen gas. How many grams of aluminum sulfate would be formed if 88.5 g H2SO4 completely reacted with aluminum?

Aluminum Reacts With Sulfuric Acid To Produce Aluminum Sulfate In Hydrogen Gas How Many Grams Of Aluminum Sulfate Would Be Formed If 885 G H2SO4 Completely Reac class=

Sagot :

Explanation:

Aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas according to the following equation.

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

We have to find the mass of Al₂(SO₄)₃ that would be formed if 88.5 g of H₂SO₄ completely reacted with Al.

First we have to convert the mass of sulfuric acid into moles using its molar mass.

molar mass of H₂SO₄ = 98.08 g/mol

moles of H₂SO₄ = 88.5 g * 1 mol/(98.08 g)

moles of H₂SO₄ = 0.902 moles

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

Now, if we pay attention to the coefficients of the reaction we will see that 2 moles of Al will react with 3 moles of H₂SO₄ to produce 1 mol of Al₂(SO₄)₃ and 3 moles of H₂. So the molar ratio between H₂SO₄ and Al₂(SO₄)₃ is 3 to 1. We can use that relationship to find the number of moles of Al₂(SO₄)₃ that can be produced when 0.902 moles of H₂SO₄ reacted with excess Al.

3 moles of H₂SO₄ : 1 mol of Al₂(SO₄)₃ mole ratio

moles of Al₂(SO₄)₃ = 0.902 moles of H₂SO₄ * 1 mol of Al₂(SO₄)₃/(3 moles of H₂SO₄)

moles of Al₂(SO₄)₃ = 0.300 moles

And finally we can convert those moles back to grams using the molar mass of Al₂(SO₄)₃.

molar mass of Al₂(SO₄)₃ = 342.15 g/mol

mass of Al₂(SO₄)₃ = 0.300 moles * 342.15 g/mol

mass of Al₂(SO₄)₃ = 102.6 g

Answer: b. 103 g

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