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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.

Respuesta :

scme1702
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
taskmasters
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 

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