Chapter U4, Problem 14STP

Interpretation Introduction

Interpretation:

The correct balanced neutral reaction needs to be identified.

Concept introduction:

A chemical equation is a symbolic depiction of a chemical reaction in the form of formulae and symbols, wherein the reactant units are given on the left side and the product units on the right side. And in the balanced chemical equation, the number components present in reactant and product must be same.

Answer to Problem 14STP

Option (D) is correct.

Explanation of Solution

Reason for correct options:

The option (D) can be written as:

${\text{2 H}}_3{\text{PO}}_4\text{ + 3 Mg}{\left(\text{OH}\right)}_2\to {\text{ Mg}}_3{\left({\text{PO}}_4\right)}_2{\text{ + 6 H}}_2\text{O}$

All the atoms are balanced thus, it is a balanced chemical equation.

So, option (D) is correct.

Conclusion

Reason for incorrect option/s:

The given equation for option (A) is:

${\text{2 HNO}}_3{\text{ + CaOH}}_2\to {\text{ 2 H}}_2\text{O + Ca}{\left({\text{NO}}_3\right)}_2$

In the above reaction the equation is not balanced because the oxygen is not balanced on both sides Thus,

The option (A) can be correctly written as:

${\text{2 HNO}}_3{\text{ + Ca(OH)}}_2\to {\text{ 2 H}}_2\text{O + Ca}{\left({\text{NO}}_3\right)}_2$

The given equation for option (B) is:

${\text{H}}_2{\text{SO}}_4\text{ + KOH }\to {\text{ 2 H}}_2{\text{O + K}}_2{\text{SO}}_4$

In the above reaction the equation is not balanced because the number of potassium (K) is not balanced on both sides thus,

The option (B) can be correctly written as:

${\text{H}}_2{\text{SO}}_4\text{ + 2 KOH }\to {\text{ 2 H}}_2{\text{O + K}}_2{\text{SO}}_4$

The given equation for option (C) is:

${\text{NH}}_4{\text{OH + HC}}_2{\text{H}}_3{\text{O}}_2\to {\text{ NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2{\text{ + 2H}}_2\text{O}$

In the above reaction the equation is not balanced because the number of hydrogen atoms is not balanced on both sides thus,

The option (C) can be correctly written as:

${\text{NH}}_4{\text{OH + HC}}_2{\text{H}}_3{\text{O}}_2\to {\text{ NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2{\text{ + H}}_2\text{O}$