cyanureeves
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chromium trioxide problem
gentlemen will someone help me? h2s04+na2cr4→cro3+naso4+h2o i added 8ml concentrated sulfuric acid to 10 grams of sodium dichromate to
make the trioxide for plating solution.how do i go about figuring right amount of sulfuric acid to get the maximum chromium trioxide? gram to ml is
confusing. so is density and mm. most problems i googled deal with % of concentrate per litre.and will the cro3 precipitate or will the naso4? i just
roughly estimated g/cm ratio of both h2so4 and na2cr4.
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bbartlog
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Sodium dichromate is not Na2Cr4, but Na2Cr2O7.
Assuming that the reaction is
H2SO4 + Na2Cr2O7 → 2CrO3 + Na2SO4 + H2O,
you need equimolar quantities of H2SO4 and sodium dichromate. Using a density of 1.83g/cc for your concentrated H2SO4, it would seem you added 14.6g
(=1.83g/cc x 8cc) (~145mmol, assuming 97% acid) or so of acid to 10g (34mmol) of the dichromate (assuming it it the dihydrate). So for the
stoichiometry to be balanced, you would need 42.6g (=0.145 moles x 298g/mol) of the dichromate, not ten.
However, I expect that if you use that much, your products will not be in solution; there will only be about 3-4g of water for 65g of salts, i.e. you
would end up with a cake even if your reaction went to completion. For this reason, I would suggest using more water. I guess it depends on your
separation strategy, though; maybe leaching such a cake with ice cold water would be a reasonable way to try to separate the CrO3 from most of the
sulfate.
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cyanureeves
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dichromate
thank you much bbartlog. and yes youre right about the cake and the chromate. i guess i over shot a bit. i mean back in 1977 when i said. "i dont need
math".i will try the cold water of course.
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