Question

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?

Answer 1

When 88.5 g of sulfuric acid reacts with aluminum, it would produce 102.9 g of aluminum sulfate, as calculated using the balanced chemical equation and moles-to-grams conversion.

Step-by-step explanation:

To calculate the amount of aluminum sulfate produced from 88.5 g of sulfuric acid (H2SO4), we first need to write the balanced chemical equation:

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

Now, we use molar masses to find the amount of aluminum sulfate produced. The molar mass of H2SO4 is 98.079 g/mol, and the molar mass of Al2(SO4)3 is 342.15 g/mol. We calculate moles of H2SO4:

Moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4

Moles of H2SO4 = 88.5 g / 98.079 g/mol = 0.902 moles of H2SO4

From the stoichiometry of the balanced equation, 1 mole of H2SO4 reacts to produce 1/3 mole of Al2(SO4)3. Therefore, we multiply the moles of H2SO4 by 1/3 to get moles of Al2(SO4)3:

Moles of Al2(SO4)3 = 0.902 moles of H2SO4 × (1/3)

Moles of Al2(SO4)3 = 0.3007 moles

Finally, we convert moles of Al2(SO4)3 to grams:

Mass of Al2(SO4)3 = moles of Al2(SO4)3 × molar mass of Al2(SO4)3

Mass of Al2(SO4)3 = 0.3007 moles × 342.15 g/mol = 102.9 g of Al2(SO4)3

So, 88.5 g of H2SO4 would produce 102.9 g of aluminum sulfate.

Answer 2

Step-by-step 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