Calculate the volume of hydrogen that is formed by the interaction of 5 mol of AL with hydrochloric acid.

| February 17, 2021 | education

Calculate the volume of hydrogen that is formed by the interaction of 5 mol of AL with hydrochloric acid. How many grams of HCL is required for this reaction?

Data: νAl – amount of the substance of the reacted aluminum (νAl = 5 mol).

Const: Vm – molar volume (we consider the reaction conditions as normal and Vm = 22.4 l / mol); MHCl is the molar mass of hydrochloric acid (MHCl = 36.46 g / mol).

1) The considered reaction: 2AlCl3 (aluminum chloride) + 3H2 (hydrogen) ↑ = 2Al + 6HCl.

2) Amounts of substance:

– aluminum: νAl / νH2 = 2/3 and νH2 = 3 * νAl / 2 = 3 * 5/2 = 7.5 mol;

– hydrochloric acid: νAl / νHСl = 2/6 = 1/3 and νHСl = 3 * νAl / 1 = 3 * 5 = 15 mol.

3) The volume of hydrogen: VH2 = νH2 * Vm = 7.5 * 22.4 = 168 liters.

4) The mass of the required hydrochloric acid: mHCl = νHСl * МHСl = 15 * 36.46 = 546.9 g.

Answer: 168 liters of hydrogen were formed, hydrochloric acid took 546.9 g.



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