The volume of hydrogen released during the interaction of sodium with 200 g of hydrochloric

| June 10, 2021 | education

The volume of hydrogen released during the interaction of sodium with 200 g of hydrochloric acid solution containing 36.5% acid is equal to ..

Data: mр is the mass of the acid solution (mр = 200 g); ωНCl – the content of hydrochloric acid in the taken solution (ωНCl = 36.5% = 0.365).

Const: MHCl – molar mass of hydrochloric acid (MHCl = 36.46 g / mol, assumed MHCl ≈ 36.5 g / mol); Vm – molar volume (for n.o. Vm = 22.4 l / mol).

1) Find out the mass of hydrochloric acid in the taken solution: mНCl = mр * ωНCl = 200 * 0.365 = 73 g.

2) Considered reaction: 2NaCl (sodium chloride) + H2 (hydrogen) ↑ = 2Na (sodium) + 2HCl (hydrochloric acid).

3) Amount of hydrochloric acid substance: νHCl = mHCl / MHCl = 73 / 36.5 = 2 mol.

4) Amount of hydrogen substance: νH2 / νHCl = 1/2 and νH2 = νHCl / 2 = 2/2 = 1 mol.

5) The volume of hydrogen produced: VH2 = Vm * νH2 = 22.4 * 1 = 22.4 liters.

Answer: The volume of released hydrogen, according to the calculation, is 22.4 liters.



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