Determine the amount of hydrogen that will be released as a result of the interaction of lithium with 300 g

| January 11, 2021 | education

Determine the amount of hydrogen that will be released as a result of the interaction of lithium with 300 g of 9.8% phosphoric acid solution.

These tasks: mр is the mass of the taken phosphoric acid solution (mр = 300 g); ωН3РО4 – mass fraction of phosphoric acid (ωН3РО4 = 9.8% = 0.098).
Const: MH3PO4 – molar mass of phosphoric acid (MH3PO4 = 98 g / mol); Vm – molar volume (assumed n.v. and Vm = 22.4 l / mol).
1) The mass of phosphoric acid: mН3РО4 = ωН3РО4 * mр = 0.098 * 300 = 29.4 g.
2) Amount of phosphoric acid substance: νH3PO4 = mH3PO4 / MH3PO4 = 29.4 / 98 = 0.3 mol.
3) Reaction of hydrogen evolution: 2Li3PO4 (lithium orthophosphate) + 3H2 (hydrogen) = 6Li (lithium) + 2H3PO4 (phosphoric acid).
4) Amount of hydrogen substance: νH2 / νН3PO4 = 3/2, whence we express: νH2 = 3 * νН3PO4 / 2 = 3 * 0.3 / 2 = 0.45 mol.
5) The volume of hydrogen: VH2 = Vm * νH2 = 22.4 * 0.45 = 10.08 liters.
Answer: 10.08 liters of hydrogen were released.



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