1 / 4

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

magda
Télécharger la présentation

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 =

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related