Chapter 23, Problem 74AP

Interpretation Introduction

Interpretation:

Schemes $\text{I}$ and $\text{II}$ are to be identified that involve magnesium, and equations that are involved in the reaction are to be represented.

Concept Introduction:

In a balanced equation, the number of atomsshould be equal on both the sides.

In a balanced chemical reaction, the mass of the reactants is equal to the mass of the products.

In the case of a nonbalanced equation, choose big number in front of each atom to do balance equation.

A chemical reaction consists of the atoms or molecules that combine with each other to give a product.

Reactants are written in left side of chemicalreaction and products are written in right side of chemical reaction.

Answer to Problem 74AP

Solution: The equations of schemes $\text{I}$ and $\text{II}$ are represented as:

Scheme $\text{I}$:

$2\text{Mg}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2\text{MgO}\left(s\right)$

$\text{A}=\text{MgO}\left(s\right)$

$\text{MgO}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{MgCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

$\text{B}={\text{MgCl}}_2\left(aq\right)$

${\text{MgCl}}_2\left(aq\right)+{\text{Na}}_{\text{2}}{\text{CO}}_3\left(aq\right)\to {\text{MgCO}}_3\left(s\right)+2\text{NaCl}\left(aq\right)$

$\text{C}={\text{MgCO}}_3\left(s\right)$

${\text{MgCO}}_3\left(s\right)\to \text{MgO}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$

$\text{D}=\text{MgO}\left(s\right)$, $\text{E}={\text{CO}}_2\left(g\right)$

${\text{CO}}_{\text{2}}\left(g\right)+\text{Ca}{\left(\text{OH}\right)}_2\left(aq\right)\to {\text{CaCO}}_3\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

$\text{F}={\text{CaCO}}_{\text{3}}\left(s\right)$

Scheme $\text{II}$:

$\text{Mg}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{MgSO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

$\text{G}={\text{MgSO}}_4\left(aq\right)$

${\text{MgSO}}_4\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)+{\text{Na}}_{\text{2}}{\text{SO}}_4\left(aq\right)$

$\text{H}=\text{Mg}{\left(\text{OH}\right)}_2\left(s\right)$

$\text{Mg}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{HNO}}_3\left(aq\right)\to \text{Mg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

$\text{I}=\text{Mg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(s\right)$

$\text{2Mg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(s\right)\to +2\text{MgO}\left(s\right)+4{\text{NO}}_2\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)$

${\text{NO}}_{\text{2}}$ is considered as brown gas

Explanation of Solution

Scheme $\text{I}$:

When solid magnesium burns in oxygen, it gives the white solid $\text{MgO}$ as

$2\text{Mg}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2\text{MgO}\left(s\right)$.

When solid $\text{MgO}$ dissolves in aqueous $\text{HCl}$, it gives a colorless ${\text{MgCl}}_2$ solution and water as

$\text{MgO}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{MgCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

When the ${\text{MgCl}}_2$ solution reacts with aqueous ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$, it gives a white precipitate ${\text{MgCO}}_3$ and aqueous $\text{NaCl}$ as

${\text{MgCl}}_2\left(aq\right)+{\text{Na}}_{\text{2}}{\text{CO}}_3\left(aq\right)\to {\text{MgCO}}_3\left(s\right)+2\text{NaCl}\left(aq\right)$.

When ${\text{MgCO}}_3$ is heated, it decomposes into solid $\text{MgO}$ and colorless carbon dioxide gas evolved as

${\text{MgCO}}_3\left(s\right)\to \text{MgO}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

When carbon dioxide gas is passed through lime water, it gives a white precipitate ${\text{CaCO}}_{\text{3}}$ and water as

${\text{CO}}_{\text{2}}\left(g\right)+\text{Ca}{\left(\text{OH}\right)}_2\left(aq\right)\to {\text{CaCO}}_3\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Scheme $\text{II}$:

When solid magnesium reacts with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, it gives a colorless ${\text{MgSO}}_4$ solution and hydrogen gas as

$\text{Mg}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{MgSO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$.

When ${\text{MgSO}}_4$ solution reacts with aqueous $\text{NaOH}$, it gives a white precipitate $\text{Mg}{\left(\text{OH}\right)}_2$ as

${\text{MgSO}}_4\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Mg}{\left(\text{OH}\right)}_2\left(s\right)+{\text{Na}}_{\text{2}}{\text{SO}}_4\left(aq\right)$.

When $\text{Mg}{\left(\text{OH}\right)}_2$ dissolves in aqueous ${\text{HNO}}_{\text{3}}$, it gives a colorless solution of $\text{Mg}{\left({\text{NO}}_3\right)}_2$ and water as

$\text{Mg}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{HNO}}_3\left(aq\right)\to \text{Mg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

When $\text{Mg}{\left({\text{NO}}_3\right)}_2$ is slowly evaporated, it gives a white solid $\text{MgO}$ and the brown gas ${\text{NO}}_{\text{2}}$ as

$\text{2Mg}{\left({\text{NO}}_{\text{3}}\right)}_2\left(s\right)\to +2\text{MgO}\left(s\right)+4{\text{NO}}_2\left(g\right)+{\text{O}}_{\text{2}}\left(g\right).$

Conclusion