The following unbalanced equation shows one of the steps in the process of reducing iron ore to iron metal.Fe2O3(s) + CO(g) → Fe(s) + CO2(g)If you begin with 297 grams of iron(III) oxide and an excess of CO, what is the maximum number of grams of Fe that can be obtained?
Hi Megan, I am french (Boulogne sur mer 62200 - FRANCE) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) M(Fe2O3) 2M(Fe) + 3M(O) M(Fe2O3) (2 x 55,85) + (3 x 16) M(Fe2O3) 159,7 g/mole M(Fe) 55,85 g/mole 2M(Fe) (2 x 55,85) 111,7 g . . . according to the equation : 159,7 g of Fe2O3 gives 111,7 g of Fe . . . then : 297 g of Fe2O3 gives 297 x 111,7 / 159,7 207,7 g of Fe I hope to have answered your question. .
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Hi Megan, I am french (Boulogne sur mer 62200 - FRANCE) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) M(Fe2O3) 2M(Fe) + 3M(O) M(Fe2O3) (2 x 55,85) + (3 x 16) M(Fe2O3) 159,7 g/mole M(Fe) 55,85 g/mole 2M(Fe) (2 x 55,85) 111,7 g . . . according to the equation : 159,7 g of Fe2O3 gives 111,7 g of Fe . . . then : 297 g of Fe2O3 gives 297 x 111,7 / 159,7 207,7 g of Fe I hope to have answered your question. .
