Limiting Reactants

Question

Suppose 49.4 g of hydrobromic acid is mixed with 30. g of sodium hydroxide. What is the maximum mass of sodium bromide that could be produced by the chemical reaction?

Answer 1

HBr + NaOH ===> NaBr + H2O

moles HBr present = 49.4 g x 1 mole HBr/80.9 g = 0.611 moles HBr

moles NaOH = 30. g x 1 mole NaOH/40 g = 0.75 moles NaOH

Limiting reactant is HBr since mole ratio is 1:1 and there are fewer moles of HBr than NaOH

moles of NaBr that can be produced = 0.611 moles since mole ratio of NaBr to HBr is 1:1

mass of NaBr produced = 0.611 moles x 103 g/mole = 62.9 g (to 3 significant figures)