Zn+2HCl⇒ZnCl₂+H₂

Assuming you start with 10.0 g of zinc and 10.0 g hydrochloric acid, identify the limiting reagent and determine what mass of the other will remain after the reaction runs to completion.

The balanced chemical reaction for the substances given would be as follows:

Zn + 2HCl = ZnCl2 + H2

We are given the amounts of the reactants used in the reaction. We use these amounts to determine which is the limiting and excess reactant. We do as follows:

10 g Zn (1 mol / 65.38 g) = 0.1530 mol

10 g HCl (1 mol / 36.46 g) = 0.2743 mol

From the the stoichiometric ratio which is 1 is to 2, the limiting reactant would be hydrochloric acid and the excess would be zinc metal.

Mass of zinc that remains = 0.1530 - (0.2743 / 2) = 0.0159 g Zn

The equation shows that one mole of Zn requires 2 moles of HCl. We now calculate the moles of each specie:

Zn = 10/65 = 0.154 mole

HCl = 10/36.5 = 0.274 mole

Molar ratio present:

1: 1.78; thus, the HCl is the limiting reactant

Moles of Zn reacted = 0.274/2 = 0.137

Moles of Zn remaining = 0.154 - 0.137

= 0.017 moles

Mass of Zn left = 0.017 x 65

= 1.105 grams