4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g) A sample of 38.5 g MnO2 is added to a solution containing 50.3 g HCl. What is the limiting reactant?

Answer 1

4HCl(aq) + MnO₂(s) ==> MnCl₂(aq) + 2H₂O(l) + Cl₂(g) ... balanced equation

An easy way to find the limiting reactant is to simply divide the moles of each reactant by the respective coefficient in the balanced equation. Thus...

for HCl: 50.3 g HCl x 1 mol HCl/36.5 g = 1.38 mols HCl (÷4 -> 0.345)

for MnO₂: 38.5 g MnO₂ x 1 mol MnO₂ / 86.9 g = 0.443 mol MnO₂ (÷1 -> 0.443)

Since 0.345 is less than 0.443, HCl is LIMITING

Of course, you could use the 1.38 mols of HCl and the 0.443 mols MnO₂ and calculate how much product (either MnCl₂, H₂O or Cl₂) is formed, and whichever produces the least amount of product would be limiting, but this first method is quicker and simpler.