Stoichiometry question

Consider the reaction: HBrO3(aq) + 5HBr(aq) -> 3H2O(l) + 3Br2(l)

What mass of H2O is produced when 1.25 dm3 of 0.45 mol dm-3 HBr(aq) are reacted?

well let's approach this problem logically for a second now.... well considering that we have 3 moles of water mixed with 3 moles of bromine, you can factor the 2 and cancel out the cross product of the square of solutions and get 6.0804g



in case that didn't make that much sense to you, ill do it the way that most novices do these types of problems... well i mean, you probably aren't familiar with it... oh whatever.

alright so we know that 1.25dm3 = 1.25L sorry i just hate dm3...

now it's just a simply conversion bridge:
(0.45mol/L)*(1.25L)*(3mol H2O/5mol HBr)*(18.016g/1mol)= 6.0804g H2O

ps you have to assume that HBr is the limiting reactant in this problem for this to work.

HBrO3(aq) + 5HBr(aq) -> 3H2O(l) + 3Br2(l)

1.2L x 0.45M = 0.54moles HBr
0.54moles HBr (3H2O/5H2O) x 18g/mole = 5.83g

mol of HBr = 1.25 dm3 X 0.45 mol dm-3 = 0.56 mol
mol H2O= 0.56 mol HBr X ( 3molH2O / 5Mol HBr)= 0.34 mol
m *** of H2O= .34mol X 18.0g/mol= 6.1g