9 November, 23:26

# How many grams of aluminum sulfate form when 3.90 grams of aluminum is placed in 13.65 grams of sulfuric acid?

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1. 9 November, 23:32
0
Mass = 17.12 g

Explanation:

Given dа ta:

Mass of Al = 3.90 g

Mass of H₂SO₄ = 13.65

Mass of aluminium sulfate = ?

Solution:

Chemical equation:

3H₂SO₄ + 2Al → Al₂ (SO₄) ₃ + 3H₂

Now we will calculate the number of moles of each reactant.

Moles of H₂SO₄:

Number of moles = mass / molar mass

Molar mass of H₂SO₄ = 98.079 g/mol

Number of moles = 13.65 g / 98.079 g/mol

Number of moles = 0.14 mol

Moles of Al:

Number of moles = mass / molar mass

molar mass of Al = 27 g/mol

Number of moles = 3.90 g / 27 g/mol

Number of moles = 0.14 mol

Now we will compare the moles of aluminium sulfate with sulfuric acid and aluminium.

H₂SO₄ : Al₂ (SO₄) ₃

3 : 1

0.14 : 1/3*0.14 = 0.05

Al : Al₂ (SO₄) ₃

2 : 1

0.14 : 1/2*0.14 = 0.07

The number of moles of aluminium sulfate produced by sulfuric acid are less so it will limiting reactant and limit the amount of aluminium sulfate.

Mass of aluminium sulfate:

Mass = number of moles * molar mass

molar mass of aluminium sulfate = 342.15 g/mol

Mass = 0.05 mol * 342.15 g/mol

Mass = 17.12 g