2AL + 6HCL = 2ALCL3 + H2 calculate the volume of gas evolved and the mass of the salt obtained

2AL + 6HCL = 2ALCL3 + H2 calculate the volume of gas evolved and the mass of the salt obtained in the reaction if 5.4 g of aluminum metal and 21 g of acid were used for it

1. Reaction equation:

2Al + 6HCl = 2AlCl3 + 3H2 ↑.

2. Let’s calculate the chemical amount of aluminum:

n (Al) = m (Al): M (Al) = 5.4: 27 = 0.2 mol.

3. Find the amount of hydrochloric acid:

n (HCl) = m (HCl): M (HCl) = 21: 36.5 = 0.5753 mol.

4. In a slight deficiency, hydrogen chloride was taken, we determine from its data the amount of formed aluminum chloride and released hydrogen:

n (AlCl3) = n (HCl): 3 = 0.5753: 3 = 0.1918 mol;

n (H2) = n (HCl): 2 = 0.5753: 2 = 0.2877 mol.

5. Calculate the mass of salt:

m (AlCl3) = n (AlCl3) * M (AlCl3);

M (AlCl3) = 27 + 35.5 * 3 = 133.5 g / mol;

m (AlCl3) = 0.1918 * 133.5 = 25.6 g.

6. Set the volume of hydrogen:

V (H2) = n (H2) * Vm = 0.2877 * 22.4 = 6.44 liters.

Answer: 25.6 g AlCl3, 6.44 l H2.