How many grams of FeCr2O7 are needed for 44.0 grams of CO2?

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SUMMARY

To produce 44.0 grams of carbon dioxide (CO2), 135.9 grams of iron (II) dichromate (FeCr2O7) are required based on the stoichiometric relationship in the reaction 4FeCr2O7 + 8K2CO3 + O2 --> 2Fe2O3 + 8K2CrO4 + 8CO2. The molar mass of iron (II) dichromate is established as 271.85 g/mol, leading to the calculation that 0.5 moles of FeCr2O7 yield 1 mole of CO2. The discussion emphasizes the use of proportions for stoichiometric calculations, which is a preferred method among participants.

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Homework Statement


4FeCr2O7 + 8K2CO3 + O2 --> 2Fe2O3 + 8K2CrO4 + 8CO[sub2[/sub]

How many grams of iron (II) dichromate are required to produce 44.0 grams of carbon dioxide?


Homework Equations





The Attempt at a Solution



79.925 grams of iron (II) dichromate
 
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Too low. Show your work.

Strange reaction and strange reactant though. If you ask me, there is no such thing as iron (II) dichromate.
 
Nevermind my first answer, I misinterpreted the problem, this is my new answer:

Since it takes takes 4 moles of iron (II) dichromate to produce 8 moles of carbon dioxide, it would take 0.5 moles of iron (II) dichromate to produce 1 mole of carbon dioxide (1 mole of CO2 = 44 g). And 0.5 moles of iron (II) dichromate = 109.92 grams of iron (II) dichromate (answer).

-- I don't really know what I'm doing btw; these are questions from my pre-AP Chem summer assignment and I'm bookless and haven't done any chemistry for about 1 1/2 years.
 
Your approach is OK, just numbers are wrong. What have you used for molar mass of iron (II) dichromate?
 
molar mass of iron (II) dichromate = 271.85 g

271.85/2 = 135.93 g... nevermind my other answer, is 135.93 g correct?
 
135.9 g is OK. Sorry for the late answer.
 
molar mass 4FeCr2O7 = 1087.4 g

molar mass 8CO2 = 352.08 g

1087.4/352.08 = x/44 ~ x = 135.9 g

^ Is it possible to use proportions for these type of problems?
 

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