Question

Question 5 Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al(s) + 302 (g) → 2A1203 (s) The maximum amount of A1203 that can be produced from 2.5 g of Al and 2.5 g of O2 is 5.0 ③ 4.7 3 9.4 * 5.3 * 7.4

Answer 1

Question - 5 Geiven, balanced reaction, GAL(S) + 302 (8) ► 241203 (3) Molecular weight & aluminium = 27 g/mol 2.58 y Al = 216 mole of Al = 0.09 mole of Al Molecular weight of oxygen = (2x16) almo) = 32 g/mol :. 25 g h O2 = 2.3 mole of O2 = 0.078 mole of O2 The know, that amount of gield calculated by the reagent. by the limits limiting

Again we know that which (n) is low, it is limiting reagent. where, no no y mole of reactant sie = stoichiometric ca efficient For Ali for 2: n= sie = 0.09 4 n=0.078 sie = 3 - ) - 009 ) = 0.026 = 0.0225 So, here Al is the limiting reagent. Molecular weight of Alo, = [(2x27) + (3X16)] g/mol 102 g/mol 2 mole of Al2O3 = (2x102) = 204 almal and 4 mole of Al = (4x27) = 108 g/mol So, from the balanced reaction, are can say from 108 g Al, we get 204 8 of Al₂O3 from 215g Al, we get ( 204 x 2.5 g of AĻU, = 4.7 g y Ahoz [option- Hence, 4.7g & Al₂O3 will be produced.

Question 3 Geiven, balanced reaction, 4 NH₃ (4) + 70₂(1) 4NO₂ (2) + 6H20 (2) Molecular weight of NH3 = [14+ ( 13) almol - 17 almol . 4 mole of NH3 = (4x14) = 68 g NH3 Molecular weight of 2 = (1682) Imol = 32 g/mol i 7 mole of O₂ = (A x 32 g = 224 g 02 from the balanced reaction are know, 68 8 of NH3 consumed 224 8 4 02 28.8 g of Nit consumed (220 x 28.8) & of 2 = 94.87 of 2 ~ 94.7 of a [option ③ Hence, 9417 g foxygen consumed. question - 4 Geiven, balanced reaction C3H8 (2) + 50₂ (2) 3 0₂ (2) + 4 HO (2) so here we see, from 5 mole yor, 4 moles of ho produced. in from 215 mole ry 0₂, (4 x 2.5) moles of the produced, = 2.0 moles of tho produced. Coption - 67 Hene, 2.0 moles of tho will produced.