Homework Answers
no of moles of Ba(OH)2 = molarity*volume in L
= 5.28*10^-2*0.0283 = 0.00149424moles
Ba(OH)2(aq) + 2HBr(aq) ----------------> BaBr2(aq) + 2H2O(l)
1 mole of Ba(OH)2 react with 2 moles of HBr
0.00149424moles of Ba(OH)2 react with = 2*0.00149424/1 = 0.00298848 moles of HBr
mass of HBr = no of moles * gram molar mass
= 0.00298848*80.91 = 0.2418g
percent by mass of HBr = 0.2418*100/8.45 = 2.86%
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