Question

In one experiment you reacted 45.7 g of Aluminum with 182.8 g of Iron (111) oxide to produce Aluminum Oxide and Elemental iron (not diatomic) (10 pts each)

Answer 1

Ada O₂ + 2fe Imole zmole 2ALt Fe2O₂ 2mole Imole - 45.7g 182.8m no of molemass = 182.8 g Molar 160 g/mol Mans = 45.79 - lu mote - 27 gloole = 1.69. mole so zmole 1.69" of u Al need = Imole of Fe₂O3 a a = 1x1.69 mole of - Ferooz = 0.845 mole I so Al will be limiting reagent | 2mole of Al give = amole of the 1.69 u u u = 169 moll of te = 1.69 note x 56 grote = 94.64g of te If we got = log of te to yield = actual yield 4.00 Theoretical yield

= log x 100 co - 94.64g = 10:5676 sq Al₂O3 Molar mass of Al2O3 = 2x27+ 3x16 = 102 g/mole 102 g of Al₂O3 contain = 54 g of sl. ca a u = 549 x5 g sq 1025 = 2.64g 27g of Al contain = 1.02381023 atom - = Arogaro's No. 2.64g u = 1:023 81023x2.648 - 27g = 0./ 2001023 atom = 1.20 x 1022 atoms

(h) - It it is a required enothermic reaction the heat. if (i) If reaction required a lot energy than the activatias of reaction is quite of energy high.