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1) a) H = 2.06%, S = 32.69%, O = 65.25% b) Ca = 54.09%, O = 43.18%, H = 2.73% 2) 24 x .787 + 25 x .101 + 26 x .112 =

1 mol O 2. 2 mol MgO . x . x . 32.00 g O 2. 1 mol O 2. 40.31 g MgO. x . 1 mol MgO. 1) a) H = 2.06%, S = 32.69%, O = 65.25% b) Ca = 54.09%, O = 43.18%, H = 2.73% 2) 24 x .787 + 25 x .101 + 26 x .112 = 24.3 3) a) 98.08 g/mol, b) 759.70 g/mol

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1) a) H = 2.06%, S = 32.69%, O = 65.25% b) Ca = 54.09%, O = 43.18%, H = 2.73% 2) 24 x .787 + 25 x .101 + 26 x .112 =

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  1. 1 mol O2 2 mol MgO x x 32.00 g O2 1 mol O2 40.31 g MgO x 1 mol MgO 1) a) H = 2.06%, S = 32.69%, O = 65.25% b) Ca = 54.09%, O = 43.18%, H = 2.73% 2) 24x.787 + 25x.101 + 26x.112 = 24.3 3) a) 98.08 g/mol, b) 759.70 g/mol 4) a) 0.12 mol (= 16 g CuCl2 x 1 mol/134.45 g) b) 4091 g (= 70 mol NaCl x 58.44 g/mol) 5) there are 6.02 x 1023 particles in a mole … a) 2.41 x 1024 molecules (4 x 6.02x1023) b) 4.82 x 1024 atoms (4 x 2 x 6.02x1023) c) 2.08 x 1023 atoms (0.173 x 2 x 6.02x1023) 6) (2Mg + O2 2MgO) 142 g MgO (see below) # g MgO= 56.3 g O2

  2. 7 a) CuCl2 is simplest (not molecular - its ionic) b) Simplest and molecular c) Molecular, d) Molecular 8 a) Simplest formula is CH3 : if we assume we have 100 g: 80 g C (6.66 mol), 20 g H (19.8 mol). simplest ratio: C (6.66/6.66) = 1 mol, H (19.8/6.66) = 3.0 mol b) 30 g/mol / 15.04 g/mol = 2.0 … C2H6 9 a) 2C40H82 + 121O2 80CO2 + 82H2O, b) 12H2O + Al4C3 3CH4 + 4Al(OH)3 10a) 15064Gd 42He + 14662Sm b) 6027Co 0-1e + 6028Ni

  3. actual 15 g H2O = theoretical 23.178 g H2O = = = 41.24 g H2O 65% 23.178 g H2O 1 mol NH3 1 mol O2 6 mol H2O 6 mol H2O 18.02 g H2O 18.02 g H2O x x x x x x 17.04 g NH3 32.00 g O2 3 mol O2 4 mol NH3 1 mol H2O 1 mol H2O # g H2O= 11) Loss of product, incomplete reactions, side reactions, and impure reactants or products 12) a) O2 is limiting (see below). b) 23.18 g 20.58 g O2 26 g NH3 % yield = x 100% x 100%

  4. actual 97 g Cu2S = theoretical 125.2 g Cu2S = = = 248 g Cu2S 125.2 g Cu2S 77% 1 mol S 1 mol Cu 1 mol Cu2S 1 mol Cu2S 159.16 g Cu2S 159.16 g Cu2S x x x x x x 63.55 g Cu 32.06 g S 1 mol S 2 mol Cu 1 mol Cu2S 1 mol Cu2S # g Cu2S= 100 g Cu 13) 2Cu + S Cu2S 50 g S % yield = x 100% x 100% For more lessons, visit www.chalkbored.com

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