Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water

Question

Suppose 24. g of hydrobromic acid is mixed with 22.1 g of sodium hydroxide. Calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

Hydrobromic acid = HBr

Sodium hydroxide = NaOH

HBr(aq) + NaOH(s) ==> NaBr(aq) + H2O(l) ... balanced equation

molar mass HBr = 80.9 g/mol

molar mass NaOH = 58.4 g/ mol

molar mass NaBr = 103 g / mol

Since we are provided with the amounts of BOTH reactants, we need to find which one is present in limiting supply. One way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For HBr: 24. g x 1 mol / 80.9 g = 0.297 mols HBr

For NaOH: 22.1 g x 1 mol / 58.4 g = 0.378 mols NaOH

Since both have a coefficient of 1, the HBr is the limiting reactant

We now use mols of HBr (0.297) to find maximum mass of NaBr that can form:

0.297 mol HBr x 1 mol NaBr / mol HBr x 103 g / mol = 30.6 g NaBr

Rounding to 2 significant figures based on 2 sig,figs, in the 24. g of HBr, we have 31 g NaBr