Chapter 13, Problem 10E

(a)

To determine

Balance and complete the given acid-base reaction: $\text{HCl}+\text{Ba}{\left(\text{OH}\right)}_2\to$

Answer to Problem 10E

The balanced acid-base reaction is written as,

$\text{2HCl}+\text{Ba}{\left(\text{OH}\right)}_2\to {\text{BaCl}}_2+2{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

Given info: The given reaction is a type of acid-base reaction.

Explanation:

The reaction between acid and base gives water and salt is called acid-base reaction. The reaction between acid and a carbonate or hydrogen carbonate to give salt, water and carbon dioxide is called acid-carbonate reaction.

The given reaction is a type of acid-base reaction. The reaction is written as,

$\text{HCl}+\text{Ba}{\left(\text{OH}\right)}_2\to {\text{BaCl}}_2+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there is one chlorine atom on the reactant side and two chlorine atoms on the product side. Therefore, multiply $\text{HCl}$ with $2$. The above reaction is written as,

$\text{2HCl}+\text{Ba}{\left(\text{OH}\right)}_2\to {\text{BaCl}}_2+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there are four hydrogen atoms and two oxygen atoms on the reactant side however there are two hydrogen atoms and one oxygen atom on the product side. Therefore, multiply ${\text{H}}_{\text{2}}\text{O}$ with $2$. The above reaction is written as,

$\text{2HCl}+\text{Ba}{\left(\text{OH}\right)}_2\to {\text{BaCl}}_2+2{\text{H}}_{\text{2}}\text{O}$

Conclusion:

Therefore, the balanced acid-base reaction is written as,

$\text{2HCl}+\text{Ba}{\left(\text{OH}\right)}_2\to {\text{BaCl}}_2+2{\text{H}}_{\text{2}}\text{O}$

(b)

To determine

Balance and complete the given acid-carbonate reaction:

$\text{HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to$

Answer to Problem 10E

The balanced acid-carbonate reaction is written as,

$\text{6HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to 2{\text{AlCl}}_{\text{3}}+3{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

Explanation of Solution

Given info: The given reaction is a type of acid-carbonate reaction.

Explanation:

The reaction between acid and base gives water and salt is called acid-base reaction. The reaction between acid and a carbonate or hydrogen carbonate to give salt, water and carbon dioxide is called acid-carbonate reaction.

The given reaction is a type of acid-carbonate reaction. The reaction is written as,

$\text{HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to {\text{AlCl}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

In the above reaction, there are two aluminium atoms on the reactant side and one aluminium atom on the product side. Therefore, multiply ${\text{AlCl}}_{\text{3}}$ with $2$. The chemical reaction is written as,

$\text{HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to 2{\text{AlCl}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

In the above reaction, there is one chlorine atom on reactant side however; there are six chlorine atoms on product side. Therefore, multiply $\text{HCl}$ with $6$. Therefore, the balanced acid-carbonate reaction is written as,

$\text{6HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to 2{\text{AlCl}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

In the above reaction, there are three carbon atoms on the reactant side however; there is one carbon atom on the product. Therefore, multiply ${\text{CO}}_{\text{2}}$ with $3$. The above reaction is written as,

$\text{6HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to 2{\text{AlCl}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

In the above reaction, there are six hydrogen atoms and nine oxygen atoms on the reactant side however; there are two hydrogen atoms and seven oxygen atoms on the product side. Therefore, multiply ${\text{H}}_{\text{2}}\text{O}$ with $3$. Therefore, the balanced equation is written as,

$\text{6HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to 2{\text{AlCl}}_{\text{3}}+3{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

Conclusion:

Therefore, the balanced acid-carbonate reaction is written as,

$\text{6HCl}+{\text{Al}}_{\text{2}}{\left({\text{CO}}_{\text{3}}\right)}_{\text{3}}\to 2{\text{AlCl}}_{\text{3}}+3{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

(c)

To determine

Balance and complete the given acid-bicarbonate reaction:

${\text{H}}_3{\text{PO}}_4+{\text{LiHCO}}_{\text{3}}\to$

Answer to Problem 10E

The balanced acid-bicarbonate reaction is written as,

${\text{H}}_3{\text{PO}}_4+3{\text{LiHCO}}_{\text{3}}\to {\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}+3{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

Explanation of Solution

Given info: The given reaction is a type of acid-hydrogen carbonate reaction.

Explanation:

The reaction between acid and base gives water and salt is called acid-base reaction. The reaction between acid and a carbonate or hydrogen carbonate to give salt, water and carbon dioxide is called acid-carbonate reaction.

The given reaction is a type of acid-hydrogen carbonate reaction. The reaction is written as,

${\text{H}}_3{\text{PO}}_4+{\text{LiHCO}}_{\text{3}}\to {\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

In the above reaction, there is one lithium atom on the reactant side and three lithium atoms on the product side. Therefore, multiply ${\text{LiHCO}}_{\text{3}}$ with $3$. The chemical equation is written as,

${\text{H}}_3{\text{PO}}_4+3{\text{LiHCO}}_{\text{3}}\to {\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}$

In the above reaction, there are three carbon atoms on the reactant side and there is one carbon atom on the product side. Therefore, multiply ${\text{CO}}_{\text{2}}$ with $3$. The above reaction is written as,

${\text{H}}_3{\text{PO}}_4+3{\text{LiHCO}}_{\text{3}}\to {\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

In the above reaction, there are six hydrogen atoms and thirteen oxygen atoms on the reactant side however; there are two hydrogen atoms and eleven oxygen atoms on the product side. Therefore, multiply ${\text{H}}_{\text{2}}\text{O}$ with $3$. The balanced acid-hydrogen carbonate reaction is written as,

${\text{H}}_3{\text{PO}}_4+3{\text{LiHCO}}_{\text{3}}\to {\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}+3{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

Conclusion:

Therefore, the balanced acid-hydrogen carbonate reaction is written as,

${\text{H}}_3{\text{PO}}_4+3{\text{LiHCO}}_{\text{3}}\to {\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}+3{\text{H}}_{\text{2}}\text{O}+3{\text{CO}}_{\text{2}}$

(d)

To determine

Balance and complete the given acid-base reaction: $\text{Al}{\left(\text{OH}\right)}_{\text{3}}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to$

Answer to Problem 10E

The balanced acid-base reaction is written as,

$\text{2Al}{\left(\text{OH}\right)}_{\text{3}}+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3+6{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

Given info: The given reaction is a type of acid-base reaction.

Explanation:

The reaction between acid and base gives water and salt is called acid-base reaction. The reaction between acid and a carbonate or hydrogen carbonate to give salt, water and carbon dioxide is called acid-carbonate reaction.

The given reaction is a type of acid-base reaction. The reaction is written as,

$\text{Al}{\left(\text{OH}\right)}_{\text{3}}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there is one aluminium atom on the reactant side and two aluminium atoms on the product side. Therefore, multiply $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ with $2$. The above reaction is written as,

$\text{2Al}{\left(\text{OH}\right)}_{\text{3}}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there is one ${\text{SO}}_{\text{4}}$ unit on the reactant side and three ${\text{SO}}_{\text{4}}$ units on the product side. Therefore, multiply ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ with $3$. The above reaction is written as,

$\text{2Al}{\left(\text{OH}\right)}_{\text{3}}+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3+{\text{H}}_{\text{2}}\text{O}$

In the above reaction, there are twelve hydrogen atoms and eighteen oxygen atoms on the reactant side however there are two hydrogen atoms and thirteen oxygen atoms on the product side. Therefore, multiply ${\text{H}}_{\text{2}}\text{O}$ with $6$. The above reaction is written as,

$\text{2Al}{\left(\text{OH}\right)}_{\text{3}}+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3+6{\text{H}}_{\text{2}}\text{O}$

Conclusion:

Therefore, the balanced acid-base reaction is written as,

$\text{2Al}{\left(\text{OH}\right)}_{\text{3}}+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_3+6{\text{H}}_{\text{2}}\text{O}$