What is the volume of hydrogen released during the interaction of 154 g of aluminum with a solution

What is the volume of hydrogen released during the interaction of 154 g of aluminum with a solution of sulfuric acid, taken in sufficient quantity, and knowing that the volume fraction of the yield of hydrogen is 90%.

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 find the chemical amount of aluminum. To do this, divide its weight by the weight of 1 mole of the substance.

M Al = 27 grams / mol;

N Al = 154/27 = 5.7 mol;

Taking into account the reaction yield of 90%, the chemical amount of released hydrogen will be: 5.7 x 3/2 x 0.9 = 7.695 mol

Let’s calculate its volume.

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

V H2 = 7.695 x 22.4 = 172.37 liters;