Ch4_B_Stoich - STOICHIOMETRY STOICHIOMETRY It rests on the...

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Unformatted text preview: STOICHIOMETRY STOICHIOMETRY It rests on the principle of the conservation It conservation of matter. - the study of the quantitative aspects of chemical reactions. PROBLEM: If 454 g of NH 44NO33 PROBLEM: If 454 g of NH NO decomposes, how much N 22O and decomposes, how much N O and H22O are formed? What is the H O are formed? What is the theoretical yield of products? theoretical yield of products? STEP 1 Write the balanced chemical equation NH4NO3 ---> N2O + 2 H 2O 2 Al(s) + 3 Br2(liq) ------> Al 2Br6(s) 454 g of NH 44NO33 ---> N22O + 2 H 22O 454 g of NH NO -> N O + 2 H O 454 g of NH 44NO33 ---> N22O + 2 H 22O 454 g of NH NO -> N O + 2 H O STEP 2 Convert mass reactant STEP Convert (454 g) --> moles STEP 3 Convert moles reactant --> STEP Convert moles product 454 g • 1 mol = 5.68 mol NH4NO3 80.04 g STEP 3 Convert moles reactant STEP Convert (5.68 mol) --> moles product Relate moles NH 4NO3 to moles product expected. 1 mol NH4NO3 --> 2 mol H2O Express this relation as the STOICHIOMETRIC FACTOR. 2 mol H 2 O produced 1 mol NH 4 NO3 used Page 1 454 g of NH 44NO33 ---> N22O + 2 H 22O 454 g of NH NO -> N O + 2 H O STEP 3 Convert moles reactant (5.68 STEP Convert mol) --> moles product 5.68 mol NH 4NO3 • 2 mol H2O produced 1 mol NH4NO3 used = 11.4 mol H2O produced 454 g of NH 44NO33 ---> N22O + 2 H 22O 454 g of NH NO -> N O + 2 H O STEP 4 Convert moles product Convert (11.4 mol) --> mass product --- called the THEORETICAL YIELD --THEORETICAL 18.02 g 11.4 mol H2 O • = 204 g H2 O 1 mol ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS! GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONS CALCULATIONS Mass product Mass reactant STEP 5 How much N 2O is formed? STEP How Total mass of reactants = total mass of products Moles reactant Stoichiometric factor 454 g NH4NO3 = ___ g N 2O + 204 g H 2O Moles product 454 g of NH 44NO33 ---> N22O + 2 H 22O 454 g of NH NO -> N O + 2 H O 454 g of NH 44NO33 ---> N22O + 2 H 22O 454 g of NH NO -> N O + 2 H O STEP 6 Calculate the percent yield STEP percent STEP 6 Calculate the percent yield STEP Calculate If you isolated only 131 g of N 2O, what is the percent yield? % yield = actual yield • 1 00% theoretical yield % yield = 131 g • 1 00% = 52.4% 250. g This compares the theoretical (250. g) This theoretical and actual (131 g) yields. and actual 454 g of NH 44NO33 ---> N22O + 2 H 22O 454 g of NH NO -> N O + 2 H O mass of N 2O = 250. g PROBLEM: Using 5.00 g of PROBLEM: Using 5.00 g of H2O2,, what mass of O 2 and of H2O2 what mass of O 2 and of H2O can be obtained? H2O can be obtained? 2 H2O2(liq) ---> 2 H2 O(g) + O2(g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2O2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2 Page 2 ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Staff during the Spring '08 term at University of Central Florida.

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