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 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