BCA table lecture

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Focus on mole relationshipsFocus on mole relationships• Step 2: fill in the before line2 H 2 S + 3 O 2 ----> 2 SO 2 + 2 H 2 OBefore: 2.4 xs 0 0ChangeAfter• Assume more than enough O 2 to react• Step 3: use ratio of coefficients to determinechange2 H 2 S + 3 O 2 ----> 2 SO 2 + 2 H 2 OBefore: 2.4 xs 0 0Change –2.4 –3.6 +2.4 +2.4After• Reactants are consumed (-), products accumulate (+)Emphasize that change and finalstate are not equivalentComplete calculations on the side• Step 4: Complete the table2 H 2 S + 3 O 2 ----> 2 SO 2 + 2 H 2 OBefore: 2.4 xs 0 0Change –2.4 –3.6 +2.4 +2.4After 0 xs 2.4 2.4• In this case, desired answer is in moles• If mass is required, convert moles to grams in theusual way3.6 moles O 2×32 .0 g1mole =115 g2
Percent Yield problems• BCA approach distinguishes between whatyou start with and what reacts. Changegrams to moles to begin.• Use the BCA table to determine moles ofproducts.• Change moles of products to grams usingmolar mass.Percent Yield problems• How many grams of aluminum iodide will form(theoretical yield) when 13.5 g of aluminumreacts with excess iodine?• 13.5 g Al x 1mole26.98 g= 0.50 molesPercent Yield problems• How many grams of aluminum iodide will form(theoretical yield) when 13.5 g of aluminumreacts with excess iodine?• 2 Al + 3 I 2 ----> 2 AlI 3Before: 0.50 xs 0ChangePercent Yield problems• 2 Al + 3 I 2 ----> 2 AlI 3Before: 0.50 xs 0Change -0.50 -0.75 +0.50After 0 xs +0.50After3
Page 1: Typical Approach to StoichiometryTh
Page 5: Limiting reactant problemsLimiting

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