Limiting Reagant problem

Question

Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 6.47 g of hydrobromic acid is mixed with 5.2 g of sodium hydroxide. Calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Answer 1

HBr + NaOH ==> NaBr + H2O ... balanced equation

molar mass HBr = 80.9 g / mol

molar mas NaOH = 40.0 g / mol

molar mass NaBr = 102.9 g / mol

One easy way to identify the limiting reactant is to divided the MOLES of each reactant by the corresponding coefficient in the balanced equation. In this case, that will be 1 for each reactant.

For HBr we have 6.47 g HBr x 1 mol / 80.9 g = 0.0798 moles HBr

For NaOH we have 5.2 g NaOH x 1 mol / 40 g = 0.13 moles NaOH

Therefore, HBr is the limiting reactant and it will determine the maximum mass of NaBr produced

Max. mass NaBr = 0.0798 mols HBr x 1 mol NaBr / mol HBr x 102.9 g / mol = 8.21 g NaBr (3 sig.figs.)