Theoretically you need 2,527 g carbon.
Fe2O3 + 3 C = 3 CO + 2 Fe
3.8g C
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.
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) ==
no it wont sorry i dont know the answer but i tried it and it didnt work for it
Taking rust to be Fe2O3, you would have the following reaction:Fe2O3 + 6HCl ==> 2FeCl3 + 3H2O100 g Fe2O3 x 1 mole Fe2O3/159.7 g = 0.626 moles Fe2O3moles HCl needed = 0.626 moles Fe2O3 x 6 moles HCl/mole Fe2O3 = 3.76 moles HCl neededMass HCl needed = 3.76 moles HCl x 36.5 g/mole = 137 g HCl needed
You need 150,41 g oxygen.
160...cant quite grasp HOW though
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.
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) ==
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no it wont sorry i dont know the answer but i tried it and it didnt work for it
Taking rust to be Fe2O3, you would have the following reaction:Fe2O3 + 6HCl ==> 2FeCl3 + 3H2O100 g Fe2O3 x 1 mole Fe2O3/159.7 g = 0.626 moles Fe2O3moles HCl needed = 0.626 moles Fe2O3 x 6 moles HCl/mole Fe2O3 = 3.76 moles HCl neededMass HCl needed = 3.76 moles HCl x 36.5 g/mole = 137 g HCl needed
We need to know the number of moles of WHAT is to react with the butane to provide you with an answer.
12
65
The balanced reaction equation is 2Fe2O3 + 3C = 4Fe + 3CO2 Note the molar ratios are 2:3::4:2 Next calculate the acutal moles of Fe2O3 First using the Periodic Table Calculate the Mr ( Relative molecular mass) of Fe2O3 Fe x 2 = 56 x 2 = 112 O x 3 = 16 x 3 = 48 112 + 48 = 160 mol(Fe2O3) = 16.5 / 160 = 0.103125 Equivlanet to ;2; moles. mol(C) = 0.103125 x 3/2 = 0.1546875 equivalent to '3 moles. Mass(C) = 0.154687 x 12 = 1.85625 g (The Answer!!!!)
40 moles of LiOH
You need 150,41 g oxygen.