Question

урд 5 For Practice 7.8 Recall from Example 7.8 that mining companies extract iron from iron ore according to the following balanced equation: Fe2O3(s) + 3 CO(g) — >2 Fe(s) + 3 CO2(g) 15979 21.. In a reaction mixture containing 167 g Fe2O3 and 85.8 g CO, CO is the limiting reactant. Calculate the mass of the reactant in excess (which is Fe2O3) that remains after the reaction has gone to completion. Imol Fe2O3 1679 X issing 1.05 mol in. co 85.8, 2015 = 3. 18 mot 112

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Answer 1

Given feo₂ = 167 gram.) ( Imol = 12+16 mees Zimol = 28 gram. co=05.8gram DATE (3 mot # 3 x 28 gram. - 04 grom tezOz (5) +300092 -> 2 fecs +3.0029) I mot 3 mot 159.69 gram 04 gram Step- & Calculate the wed mass of ₂03 in Reaction 84 gram co React with = 159.69 gram te, 03. Newtg co React with = 159.69/04 grom thoz 105.8 gram co React with = 159.60 X 85.8 gram 04 Given Amount of CO 161.6 gram fe₂03 used Amount of f203 = 161.6 gram tezo? Pemain Amount of fez oz = Given Amount -wed Amirent = 167 - 161:6 - 5. u gram flzoz or in term of mof 5.4 Remain motor mas T = 0.034 mol 159.69 + 1% mi Remain mass Thank you very much. Hope this will help you.