[Hemidol]

Selenic acid, H2SeO4, is produced from elemental selenium in a two-stage reaction. First the selenium is
oxidized to selenous acid by nitric acid. Then the selenous acid is oxidized by potassium permanganate to
selenic acid. The equations are:
3 Se (s) + 4 HNO3 (l) + H2O (l) -> 3 H2SeO3 (aq) + 4 NO (g)
8 H2SeO3 (aq) + 2 KMnO4 (aq) -> 5 H2SeO4 (aq) + K2SeO3 (aq)+ 2 MnSeO3 (aq) + 3 H2O (l)

What mass of selenic acid (in kg) can be produced from 3.80 kg of Se?


Having some difficuly with this question, any help appreciated.

[Borek]

Most obvious way would be to calculate amount of H₂SeO₃ from the first equation, then amount of H₂SeO₄ from the second.

[Hemidol]

Whoops didn't observe that when I first looked at it. How's my technique here?


H₂SeO₃ = 128.976_g

3.80 kg Se * (1000_g / 1_kg) * (1 mole Se / 78.96_g) * (3 mole H₂SeO₃ / 3 mole Se) * (128.976_g / 1 mole H₂SeO₃) * (1 kg / 1000_g) = 6.21kg

Then onto equation 2...

6.21kg H₂SeO₃ * (1000_g / 1_kg) * (5 moles H₂SeO₄ / 8 mole H₂SeO₃) * (144.976_g / 1 mole H₂SeO₄) * (1 kg / 1000_g) = 562kg


Is this the correct method? That number seems quite large to me...

[Borek]

Dimensional approach is not how I do these things, so I am not sure, but it looks to me like in the second equation you forgot to convert mass of H₂SeO₃ to moles. You are off by numerical factor of 129, that is exactly omitted molar mass.

Snack for observing that the number obtained doesn't fit correct ball park

[Hemidol]

Dimensional approach is not how I do these things, so I am not sure, but it looks to me like in the second equation you forgot to convert mass of H₂SeO₃ to moles. You are off by numerical factor of 129, that is exactly omitted molar mass.

Snack for observing that the number obtained doesn't fit correct ball park

Yeah, I'm just learning this dimensional analysis technique, I enjoy using it though. And thanks for picking up on that conversion error there... I didn't notice that before. I'll rework the problem as such:


H₂SeO₃ = 128.976_g

3.80 kg Se * (1000_g / 1_kg) * (1 mole Se / 78.96_g) * (3 mole H₂SeO₃ / 3 mole Se) * (128.976_g / 1 mole H₂SeO₃) * (1 kg / 1000_g) = 6.21kg

Then onto equation 2... (reworked)

6.21kg H₂SeO₃ * (1000_g / 1_kg) * (1 mole H₂SeO₃ / 128.976_g) (5 moles H₂SeO₄ / 8 mole H₂SeO₃) * (144.976_g / 1 mole H₂SeO₄) * (1 kg / 1000_g) = 4.362734927kg


This number appears to be more so in the range of something more "realistic", don't you agree?

Thanks very much for the snack. I was very hungry indeed.

[Borek]

Watch significant digits. Otherwise oK.