please solve part B.... The formation of organic compounds by the reaction of ir
ID: 623149 • Letter: P
Question
please solve part B....
The formation of organic compounds by the reaction of iron (II) sulfide and carbonic acid is described by the following chemical equation
2FeS+H2CO3=2FeO+(CHOH)n+2s
(A) How much FeO is produced starting with 2.30 g FeS and 0.495 moles of H2CO3 if the reaction results in a 78.50% yield?
=1.475 GRAMS
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(B) If one of the carbon containing products has a molar mass of 3.00x10^2 g/mol, what is the chemical formula of product??
Explanation / Answer
A) 2.3 gm FeS means 2.3/87.91 = 0.02616 moles moles of H2CO3 = 0.495 by equation it is clear 0.02616 moles FeS gives 0.026163 moles FeO mass of FeO = 0.02616 x 71.844 = 1.8796 grams yield = 78.5 % , hence yield of FeO = (1.8796 x78.5/100) = 1.475 grams B) (CH2O)n = 300 gm/mol C=12, H=1,O=16 then CH2O = 30 30 xn = 300 n = 10 formula is C10H20O10 = (CH2O)10
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