150 g of 20% chloride acid reacted with 3 g of aluminum to determine the volume of gas that is formed.

Given:

m (HCl) = 150 g

m (Al) = 3 g

w% (HCl) = 20%

To find:

V (gas) -?

Decision:

2Al + 6HCl = 2AlCl3 + 3H2, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of hydrochloric acid in the solution:

m (HCl) = 150 g * 0.2 = 30 g

2) Find the amount of hydrochloric acid and aluminum:

n (HCl) = m: M = 30 g: 36.5 g / mol = 0.82 mol

n (Al) = m: M = 3 g: 27 g / mol = 0.11 mol

We start from a lower value to get more accurate calculations. We work with Al:

3) We compose a logical expression:

if 2 mol of Al gives 3 mol of H2,

then 0.11 mol Al will give x mol H2,

then x = 0.11 * 3: 2 = 0.165 mol.

4) Find the volume of hydrogen released during the reaction:

V (H2) = n * Vm = 0.165 mol * 22.4 l / mol = 3.696 l.

Answer: V (H2) = 3.696 l.