Question

Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 67.2 g of hydrobromic acid is mixed with 16. g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

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:

$$m(H_{2}O)=n(H_{2}O)*M(H_{2}O)=0.40mol*18g/mol=7.2g$$

Finally, the mass of water produced is equal to 7.2 grams.