g Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 57.4 g of hydrobromic acid is mixed with 41. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Zero

Explanation:

The complete reaction for this is as follows -

HBr + NaOH = NaBr + H2O

1 mole of NaOH reacts with 1 mole of HBr

m(HBr) = 57.4g and M(HBr) = 80.9g/mol

m(NaOH) = 16g and M(NaOH) = 40g/mol

Number of moles = m/M

Substituting the given values we get -

n(HBr) = 57.4 g/80.9 g/mol = 0.709

n(NaOH) = 16 g/40 g/mol = 0.4

n(H2O) = n(NaOH) = 0.40 mol

According to balanced equation

40 gram of NaOH reacts with 81 grams of HBr

41 gram of NaOH will react with 81/40 * 41 = 83.025 gram of HBr

We have only 57.4 g of HBr only hence, 0 HBr will be left.