CHEM 105 1nd Edition Lecture 4Outline of Last Lecture • The Periodic Table of the Elements • The modern periodic table contains seven horizontal rows called periods • 18 Colums called groups or families.Outline of Current Lecture • Limiting Reactants and Percent yield • Calculations Involving Limiting Reactants • Actural Yields versus Yields Current LecturePbO(s) + NH3(g) ----> Pb(s)+ N2(g)+ H2O(I)Q: How many grams of ammonia are consumed by reaction with 80.95 g of lead(II) oxide? A: NH3- 14.01 g/mol+ 3x1.008 g/mol =17.034 g/molPbO- 207.2 g/mol +16.00g/mol +223.2 g/mol80.95 g PbOx 1 mol Pbo/223.2gx 2molNH3/3 molPbOx17.034g/ 1 mol NH3 =4.119g NH3Q: How many grams of nitrogen are formed by the reaction of 112.60g of lead(II) oxide with excess ammonia?A: N2= 4.712 g112.60g PbOx 1 mol PbO/223g x1 mol N2/3 mol PbO x28.02 g/1mol N2= 4.711845g N2- P4(s) + O2(g) ----> P4O10(s) Q: How many grams of P4O10 can be prepared from 0.375g of white phosphorus (P4)?A: P4- 4x30.97 g/mol =123.88 g/molP4O10- 4x36.97 g/mol +10x16.00 g/mol =283.88 g/mol0.375 g P4x1 mol P4/123.88g x 1mol P4O10/1 mol P4 x288.88/ 1 mol P4O10 =0.859339g P4O10 Q: How many grams of oxygen are required to completely react with 0.375g of white phosphorus?A: 0.375 g P4 x1 mol P4/123.88gx 5 mol O2/1 mol P4 x 32.00g/1 mol O2= 0.48434 g O2 Q: What mass of P4O10 can be prepared by the reaction of 5.505g of white phosphorus with 6.540g of oxygen?A: 5.505g P4x1 molP4/123.88gx1 mol P4O10/1 mol P4x 283.88g/ 1 mol= 12.615g P4O106.540 g O2 x 1mol O2/32.00g x1 mol P4O10/5 mol O2x 283.88/1 mol= 11.6036g P4O10can be made. Q: What mass of hydrogen sulfide can be prepared by the reaction of 158.16 g of aluminum sulfide and 131.13 g of water?Al2S3(s) +6H2O(I) ----> 2Al(OH3)(s) +3H2S(g)A: Al2S3= 2x26.9815g/mol+ 32.065g/mol= 150.158 g/molH2S= 2x1.008 g/mol+ 32.065g/mol= 34.081 g/mol158.06Al2S3 x1mol Al2S3/150.158g x3 mol H2S/6 mol H2Ox 34.081g/ 1 mol H2S= 107.6915841 H2S 131.13g H2O x 1mol H2O/18.016g x3 mol H2S/6 mol H2O x 34.031 g/ 1mol H2S = 124.01999 107.69 g of H2S can be
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