Question

The amount of metallic zinc (Atomic weight $= 65.4$) required to react with aqueous sodium hydroxide to produce $1{\text{g}}$ of ${{\text{H}}_2}$ is:
A. $32.7{\text{g}}$
B. $98.1{\text{g}}$
C. $65.4{\text{g}}$
D. $16.3{\text{g}}$

Answer 1

Stoichiometry is the study of the quantitative aspects of chemical reactions. Chemical equations are concise representations of chemical reactions. Mole is defined as the quantity of a substance that contains the same number of ultimate particles as are present in $12{\text{g}}$ of carbon$- 12$.

Complete step by step answer:
Given that,
Atomic weight of ${\text{Zn}}$ $= 65.4{\text{g}}$
Stoichiometry deals with the numerical relationships of elements and compounds and the mathematical proportions of reactants and products in chemical transformations. Reactants appear on the left side and products appear on the right side of the equation. Coefficients are inserted to balance the equation. Subscripts tell the number of atoms of each element in a molecule. Coefficients tell the number of molecules.
One mole of an element contains $6.022 \times {10^{23}}$ particles. This absolute number is called Avogadro’s number. Mass of one mole of substance is called molar mass. Or molar mass of an element is equal to the molecular weight.
The chemical reaction behind the process is given below:
${\text{Zn}} + 2{\text{NaOH}} \to {\text{N}}{{\text{a}}_2}{\text{Zn}}{{\text{O}}_2} + {{\text{H}}_2}$
Sodium zincate
From the chemical equation, it is clear that $1{\text{mol}}$ of ${\text{Zn}}$reacts with $2$ moles of ${\text{NaOH}}$ to give $1{\text{mol}}$ of sodium zincate and $1{\text{mol}}$ of ${{\text{H}}_2}$.
The weight of $1{\text{mol}}$ is the same as that of molecular weight.
Atomic weight of ${\text{Zn}}$ $= 65.4{\text{g}}$
Atomic weight of ${{\text{H}}_2}$ $= 2{\text{g}}$
Therefore it can be said that $1{\text{mol}}$ of ${\text{Zn}}$ gives $1{\text{mol}}$ of ${{\text{H}}_2}$.
i.e. $65.4{\text{g}}$ of ${\text{Zn}}$ gives $2{\text{g}}$ of ${{\text{H}}_2}$.
We have to find the amount of ${\text{Zn}}$ required to produce $1{\text{g}}$ of ${{\text{H}}_2}$.
i.e. Amount of ${\text{Zn}}$ $= \dfrac{{65.4{\text{g}}}}{2} = 32.7{\text{g}}$

Additional information:
Mole concepts enable us to solve stoichiometric problems involving mass relations of reactants and products in chemical reactions.

Note:
Moles provide a bridge from molecular scale to real-world scale. One mole of molecules or formula units contain Avogadro number times the number of atoms or ions of each element in the compound. Each chemical equation provides information about the amount of reactants that produce products.