In a reaction mixture of NaOH 1 mole and HCl 2 moles, which is the limiting reactant?

In a reaction mixture of NaOH 1 mole and HCl 2 moles, which is the limiting reactant?

• A. The reactant which runs out first
• B. The reactant that is in excess
• C. Both equally limiting
• D. Neither

Answer: (A)

In reactions with a fixed stoichiometry, the limiting reactant is the one that is consumed first because the reaction cannot continue once that reagent is exhausted. For NaOH and HCl, the equation is NaOH + HCl → NaCl + H2O, a 1:1 ratio. You have 1 mole of NaOH and 2 moles of HCl. To use up all the NaOH you’d need only 1 mole of HCl, and you actually have 2 moles, so NaOH will be completely consumed first, leaving excess HCl. Therefore, NaOH is the limiting reactant. The other options don’t fit because the reactant in excess would be HCl, not limiting; both equally limiting would require equal moles matching the ratio; and neither would imply that a reactant runs out.

In reactions with a fixed stoichiometry, the limiting reactant is the one that is consumed first because the reaction cannot continue once that reagent is exhausted. For NaOH and HCl, the equation is NaOH + HCl → NaCl + H2O, a 1:1 ratio. You have 1 mole of NaOH and 2 moles of HCl. To use up all the NaOH you’d need only 1 mole of HCl, and you actually have 2 moles, so NaOH will be completely consumed first, leaving excess HCl. Therefore, NaOH is the limiting reactant. The other options don’t fit because the reactant in excess would be HCl, not limiting; both equally limiting would require equal moles matching the ratio; and neither would imply that a reactant runs out.