What mass of aluminum sulfate is formed by the interaction of 9.8 grams

What mass of aluminum sulfate is formed by the interaction of 9.8 grams of sulfuric acid with an excess of aluminum? How much hydrogen is released in this case?

Metallic aluminum interacts with sulfuric acid. In this case, aluminum sulfate is synthesized and bubbles of hydrogen gas are released. The reaction is described by the following equation.

2Al + 3H2SO4 = Al2 (SO4) 3 + 3 H2;

Let’s calculate the chemical amount of sulfuric acid. To do this, divide its weight by the weight of 1 mole of the substance.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; N H2SO4 = 9.8 / 98 = 0.1 mol;

The sulfuric acid will react to produce the same amount of hydrogen gas.

Let’s calculate its volume.

To this end, we multiply the chemical amount of the substance by the volume of 1 mole of gas (filling a space with a volume of 22.4 liters).

V H2 = 0.1 x 22.4 = 2.24 liters;

The chemical amount of aluminum sulfate will be 0.1 / 3 = 0.333 mol;

Determine the weight of the substance for this, multiply the amount of the substance by the molar weight of this substance.

M Al2 (SO4) 3 = 27 x 2 + (32 + 16 x 4) x 3 = 342 grams / mol;

m Al2 (SO4) 3 = 342 x 0.333 = 113.886 grams;