The reactants are reacting according to the following equation:
HBr + NaOH = NaBr + H2O
1 mole of HBr requires 1 mole of NaOH to be totally transformed into NaBr and water.
Now, let us calculate the amounts of HBr and NaOH in mols:
m(HBr) = 67.2g
m(NaOH) = 16g
M(HBr) = 80.9g/mol
M(NaOH) = 40g/mol
Following the simple relation
"n(moles)=\\frac{m(grams)}{M(g\/mol)}""n(HBr)=\\frac{67.2g}{80.9g\/mol}=0.831mol""n(NaOH)=\\frac{16g}{40g\/mol}=0.40mol"
Thus, hydrogen bromide is in a great excess and the amount of water produced will be calculated using the amount of sodium hydroxide:
n(H2O) = n(NaOH) = 0.40 mol, which is:
Finally, the mass of water produced is equal to 7.2 grams.
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