4 FeCr2O7 + 8 K2CO3 + O2 ---> 2 Fe2O3 + 8 K2CrO4 + 8 CO2
11moles
I'm assuming you're referring to the well-known oxidation-reduction reaction that produces pure iron and carbon dioxide from iron(III) oxide and carbon monoxide. The balanced equation for that reaction is:
Fe2O3 + 3CO --> 2Fe + 3CO2
Additionally, I'm assuming that you mean how many grams of CO2 are produced, since CO is a reactant, not a product. So, based on that, here's how you proceed:
You're given 48g of iron(III) oxide, so get your molar mass of that reactant, which is 56x2+16x3=160g/mol. You have 48g, so 48/160=0.3mol. The molar ratio of this reactant to carbon dioxide product is 1:3, so take that 0.3 and triple it, so you will yield 0.9mol CO2. Molar mass of carbon dioxide is 44g/mol, and 0.9x44=39.6g, which is the mass of CO2 produced.
Since carbon dioxide is a gas, it might be more practical to give a volume produced instead, and assuming standard conditions, the volume would be 0.9 moles x 22.4 liters/mole = 20.16L. Yay stoichiometry!
22
Explanation: 44Co2 (4FeCr2O7/8CO2)
Multiply 44 by 4=176 divide by 8 22 moles of FeCr2O7
136 grams
136 grams
0.249
11
17
Fe2O3 (s) + 3 CO (g) 🡪 2 Fe (s) + 3 CO2 (g) Calculate the number of grams of CO that can react with 250 g of Fe2O3. Calculate the number of grams of Fe and the number of grams of CO2 formed when 250 grams of Fe2O3 reacts.
There are several different possible reactions of Fe2O3 with CO, depending on temperature and ratio of reactants. The simplest is probably Fe2O3 + CO ==>2FeO + CO21.00 Kg x 1000 g/Kg x 1 mole Fe2O3/160 g = 6.25 moles Fe2O3 moles CO2 produced = 6.25 moles CO2 Volume CO2 at STP = 6.25 moles x 22.4 L/mole = 140 Liters
1,4 moles of CO are produced.
Fe2O3 + 3CO -----> 2Fe + 3CO2
17
Fe2O3 (s) + 3 CO (g) 🡪 2 Fe (s) + 3 CO2 (g) Calculate the number of grams of CO that can react with 250 g of Fe2O3. Calculate the number of grams of Fe and the number of grams of CO2 formed when 250 grams of Fe2O3 reacts.
There are several different possible reactions of Fe2O3 with CO, depending on temperature and ratio of reactants. The simplest is probably Fe2O3 + CO ==>2FeO + CO21.00 Kg x 1000 g/Kg x 1 mole Fe2O3/160 g = 6.25 moles Fe2O3 moles CO2 produced = 6.25 moles CO2 Volume CO2 at STP = 6.25 moles x 22.4 L/mole = 140 Liters
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160...cant quite grasp HOW though
1,4 moles of CO are produced.
Fe2O3 + 3CO -----> 2Fe + 3CO2
It is 48 miles according to Google Maps.
Fe2O3 (s) + 3CO (g) -----> 2Fe (s) + 3CO2 (g) This is just the balanced chemical equation. It says nothing about the actual method and reaction conditions for performing the reaction. The mixture would most likely need to be heated signficantly. Why worry about dealing with carbon monoxide, when simple carbon will do? Again, significant heating of the mixture is required. 2Fe2O3 (s) + 3C (s) -----> 4Fe (s) + 3CO2 (g)
mass / molar mass molar mass Fe2O3 = 159.69 g/mol mass Fe2)3 = 4.00 kg = 4000 g moles = 4000 g / 159.69 g/mol = 25.05 moles Fe2O3 The balanced equation tells you that 1 mole Fe2O3 requires 3 moles CO to react So 25.05 moles needs (3 x 25.05) moles CO = 75.15 moles Co is needed to react 4.00 kg Fe2O3 = 75.2 mol (3 sig figs) b) The equation tells you that 1 moles Fe2O3 reacts to form 2 moles Fe So 25.05 moles will form (2 x 25.05) mol Fe moles Fe formed = 50.10 moles = 50.1 mol (3 sig figs) The equation tells you 1 mole Fe2O3 reacts to form 3 moles CO2 So 25.05 mol Fe2O3 will form (3 x 25.05) mol CO2 = 75.15 moles CO2 = 75.2 mol (3 sig figs) ==
the answer you seek is , the limiting reagent is Co, and 8.0mol Fe will be formed!
no it wont sorry i dont know the answer but i tried it and it didnt work for it