1. 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 4) a) 0.12 mol (= 16 g CuCl 2 x 1 mol/134.45 g) b) 4091 g (= 70 mol NaCl x 58.44 g/mol) 5) there are 6.02 x 10 23 particles in a mole … a) 2.41 x 10 24 molecules (4 x 6.02x10 23 ) b) 4.82 x 10 24 atoms (4 x 2 x 6.02x10 23 ) c) 2.08 x 10 23 atoms (0.173 x 2 x 6.02x10 23 ) 6) (2Mg + O 2 2MgO) 142 g MgO (see below) 2 mol MgO 1 mol O 2 x # g MgO=56.3 g O 2 40.31 g MgO 1 mol MgO x 1 mol O 2 32.00 g O 2 x

2. 7 a) CuCl 2 is simplest (not molecular - its ionic) b) Simplest and molecular c) Molecular, d) Molecular 8 a) Simplest formula is CH 3 : 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 … C 2 H 6 9 a) 2C 40 H 82 + 121O 2 80CO 2 + 82H 2 O, b) 12H 2 O + Al 4 C 3 3CH 4 + 4Al(OH) 3 10a) 150 64 Gd 4 2 He + 146 62 Sm b) 60 27 Co 0 -1 e + 60 28 Ni

3. 11) Loss of product, incomplete reactions, side reactions, and impure reactants or products 12)a) O 2 is limiting (see below). b) 23.18 g 6 mol H 2 O 4 mol NH 3 x 26 g NH 3 41.24 g H 2 O= 18.02 g H 2 O 1 mol H 2 O x 1 mol NH 3 17.04 g NH 3 x 6 mol H 2 O 3 mol O 2 x # g H 2 O= 20.58 g O 2 23.178 g H 2 O= 18.02 g H 2 O 1 mol H 2 O x 1 mol O 2 32.00 g O 2 x 15 g H 2 O 23.178 g H 2 O = % yield = x 100%65%= actual theoretical x 100%

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