Question

You must show work in order to receive credit. Make sure to include units and indicate your final answer by putting a box around your final answer. If you do not box in a response, I will count it as no final answer and you will lose points! 1. Mining companies use this reaction to obtain iron from iron ore: Fe2O3(s) + 3 CO(g) + 2 Fe(s) + 3 CO2(g) The reaction of 167 g Fe2O, with 85.8 g CO produces 72.3 g Fe. i. Determine the limiting reactant. (3pts.) 1619 Fe₂O3 1 mol Fe₂O₃ = 1.045 mol Fe₂O3 159.69 g FeO3 I limiting reacten 44.01 g CO - 85.8 g 0 x 1 mol co = 1.95 mol co Fe₂O3 ii. Determine the theoretical yield. (3pts.) 1679 Fe₂O₃ x 1 mol Fe2O3 x 55.845 g Fe x - 3 x 159.69 g Fezes Imol Fe 2 mol Fe 116.8 1 mol FeO, Fell iii. Determine the percent yield. (3pts.) ractual yield Ir 10020 I the cretical yield] x 100 1.045 mol Fe₂O3 x 159.69gFe 203 Imol Fel3

Answer 1

159.600 0 Fe2O3(5) + 3c0c9) ► 2Fe(s) + 3C029) 159.68829 3X28-0101g 2x 55-8459 1 159.68829 Fe2O3 (5) reacts with 3x28.01019 cowg) 1679. 12. 2 2 3X28.0101 X167,, 159.6882) = 87.8779gcocg) But cocg taken = 85.8 g. . The limiting reactant i's &9) (Ans) ") 3X28.01019.cogg) can produce 2x55.8459 Fe(s) 85.8 g cocg), cocg) ».. og 2x55.845 X85-8999 = 114.0429 Fe(s) :- The theoretical yield = 114.0429 Fe(s) Cans) iii) Percent yield = - Actual yield x 100% Theoretical field _172.3 X100) y 114.042 = 63.3987 7 63.40%. (Ans) 3 x 28.0101