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scoundrel [369]
3 years ago
9

Hydrogen can be obtained economically as a byproduct in the electrolysis of

Chemistry
2 answers:
4vir4ik [10]3 years ago
8 0
<span>Hydrogen can be obtained economically as a byproduct in the electrolysis of "brine".
</span>
A solution of sodium chloride (NaCl)and water (H2O) refers to the brine.The procedure of electrolysis includes utilizing an electric current to achieve a synthetic change and make new chemicals. The electrolysis of brine is a huge scale process used to make chlorine from salt, so  three important chemicals, NaOH, Cl2, H2, can be gotten by electrolyzing brine.
Ilia_Sergeevich [38]3 years ago
3 0
<span>There are too many popular methods of production of hydrogen are there. But it is one of the simplest and easy and understandable economic process is electrolysis of water. Since water contains hydrogen and oxygen gases , it has to be separated by electrolysis method. In electrolysis method Hydrogen ions and oxygen ions are separated by the electrolyzer and catched up the cathode and anode electrodes respectively. Since hydrogen ion is positively charged and attracted by the cathode and oxygen ion is negatively charged attracted by the anode. Here the following reactions are undergoing in electrolysis process as follows : 2H^2^0 ----> 2H^2 + O^2</span>
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To answer this question, use the hydrazine–nitrogen tetroxide rocket fuel reaction: 2N2H4+N2O4→3N2+4H2O You are given 96 g of N2
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Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume
Kisachek [45]

Answer:

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

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the total volume of the mixture is equal to:

Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL

the moles of each species in the mix equals:

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KI = 0.05 * 0.025 = 0.00125 moles

H2O2 = 1.02 * 0.01 = 0.0102 moles

the following equation shows the reaction between I2 and S2O32:

I2 + S2O32 = 2I- + S4O62-

The same way:

2I- + 2H+ + H2O2 = I2 + 2H2O

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

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