Chapter 3, Problem 3.86P

Interpretation Introduction

Interpretation:

The maximum amount(mol) and mass(g) of ${\text{H}}_2\text{S}$ formed by reaction of $158\text{ g}$ aluminium sulfide and $131\text{ g}$ water is to be calculated. Also the mass(g) of the excess reactant remaining is to be calculated.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Amount(mol) of excess reactant is reactant left after the formation of maximum amount(mol) of products.

The formula to calculate moles is as follows:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molecular mass}}$        (1)