Calculate the amount of aluminum sulfate produced from 10.0 grams of aluminum

Answer 1

To begin, we need to consider the following chemical equation:

2 Al + 3 H₂SO₄ --> Al₂(SO₄)₃ + 3 H₂

Using the information give, which is 10.0 g aluminum, we need to find a mole relationship between aluminum (Al) and Aluminum Sulfate.

First, convert 10.0 g Al to moles using molar mass from periodic table, which is 27 g/mol

moles Al = mass/molar mass

moles Al = 10.0g/ 27 g/mol = 0.370 moles Al

Look at the balanced equation which will give you mole relationships between Al and Aluminum sulfate. That ratio is 2 moles aluminum for every 1 mole Al₂(SO₄)₃

Next, find moles of Al₂(SO₄)₃

0.370 mol Al x 1 mol Al₂(SO₄)₃ / 2 mol Al = 0.185 mol Al₂(SO₄)₃

Finally, convert the moles of Al₂(SO₄)₃ using molar mass to grams of Al₂(SO₄)₃

Molar mass of Al₂(SO₄)₃ = (2x27) + (4x32.07 ) + (12x 16.00) = 342 g/mol

Grams Al₂(SO₄)₃ = molar mass x mass

grams Al₂(SO₄)₃ = 342 x 0.185 = 63.27 g Al₂(SO₄)₃