What volume of hydrogen will be released under normal conditions if aluminum weighing 10.8 g

What volume of hydrogen will be released under normal conditions if aluminum weighing 10.8 g is dissolved in an excess of hydrochloric acid.

Given:

m (Al) = 10.8 g;

To find:

V (H2) -?

Solution:

1) We compose the reaction equation corresponding to the condition of the problem:

2Al + 6HCl = 2AlCl3 + 3H2, – aluminum is an amphoteric metal, which means it will easily react with acids and bases.

2) Find the amount of aluminum in 10.8 grams of metal:

n (Al) = m: M = 10.8 g: 27 g / mol = 0.4 mol;

3) To find the amount of the resulting hydrogen, we compose a logical expression:

if 2 mol of Al gives 3 mol of H2 during the reaction,

then 0.4 mol of Al will give in the reaction x mol H2,

then x = 0.4 * 3: 2 = 0.6 mol.

4) Knowing the amount of the substance, we find the volume of hydrogen released as a result of the reaction:

V (H2) = n * Vm = 0.6 mol * 22.4 L / mol = 13.44 L;

Answer: V (H2) = 13.44 liters.