Preparation of Alum [KAI(SO4) 12H20] 1. Calculate the % composition of SO2. 32.0
ID: 564658 • Letter: P
Question
Preparation of Alum [KAI(SO4) 12H20] 1. Calculate the % composition of SO2. 32.0142( llo DO) :l4.UT 32.01050,057 (4.01g SO2 2. What is the theoretical yield of oxygen in grams if one mole of H20 is decomposed as : H2O H2 + ½ O2 3. If the actual yield in exercise 2 is 10 g of oxygen, what is the percentage yield? Copper (II) oxide is prepared by the reaction: 2Cu + O2 2Cuo, what is the limiting reagent if Cuo is prepared from 150 g of Cu and 50 g of O,? 4. 5. What is the maximum amount of CuO that can be prepared using the amounts of starting materials given in problem 4?Explanation / Answer
1. molar mass of SO2 = 32 + 16*2 = 64g/mole
S% = 32*100/64 = 50%
O% = 2*16*100/64 = 50%
2. H2O ---------> H2 + 1/2O2
1 mole of H2O decomposes to gives 1/2 or 0.5 moles of O2
the theoretical yield of O2 = no of moles * gram molar mass
= 0.5*32 = 16g
3. actual yield of O2 = 10g
percentage yiled = actual yield *100/ theoretical yield
= 10*100/16 = 62.5%
4. 2Cu + O2 ---------> 2CuO
no of moles of Cu = W/G.A.Wt
= 150/63.5 = 2.362 moles
no of moles of O2 = W/G.M.Wt
= 50/32 = 1.56moles
1 mole of O2 react with 2 moles of Cu
1.56 moles of O2 react with = 2*1.56/1 = 3.12 moles of Cu
Cu is limiting reagent
5.2Cu + O2 ---------> 2CuO
2 moles of Cu react with O2 to gives 2 moles of CuO
2.362 moles of Cu react with O2 to gives = 2*2.362/2 = 2.362 moles of CuO
mass of CuO = no of moles * gram molar mass
= 2.362*79.5 = 187.78g of CuO