Chapter 20, Problem 53A

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction, $Mn{O}_2\left(s\right)+HCl\left(aq\right)\to MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)+H_{2}O\left(l\right)$ is to be determined.

Concept Introduction: The oxidation number method involves the balancing of species by checking their oxidation numbers. And, correspondingly balancing all the atoms in the given reaction. If needed, water is also added to balance the equation.

Answer to Problem 53A

The balanced chemical equation is,

  $2Mn{O}_2\left(s\right)+8HCl\left(aq\right)\to \text{ 2}MnC{l}_2\left(aq\right)+2C{l}_2\left(g\right)+4H_{2}O\left(l\right)$

Explanation of Solution

The given reaction,

  $Mn{O}_2\left(s\right)+HCl\left(aq\right)\to MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)+H_{2}O\left(l\right)$

The oxidation number of $M{n}^{4+}$ in $Mn{O}_2$ changes to $+2$ in $MnC{l}_2$ by the gain of electrons, and reduction. And the oxidation number of $C{l}^{-}$ in $HCl$ changes from -1 to zero in $C{l}_2$ . The electron count is in the balanced form. Now, the count of chemical species is not equal. The count of oxygen can be balanced by the multiplication of $H_{2}O$ by 4 and the hydrogen count can be balanced by adding 8 to $HCl$ . The chlorine count can be balanced by multiplying $C{l}_2$ , $MnC{l}_2$ , and $Mn{O}_2$ by 2.

So, the balanced chemical equation becomes,

  $2Mn{O}_2\left(s\right)+8HCl\left(aq\right)\to \text{ 2}MnC{l}_2\left(aq\right)+2C{l}_2\left(g\right)+4H_{2}O\left(l\right)$

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction, $Cu\left(s\right)+HN{O}_3\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$ is to be determined.

Concept Introduction: The oxidation number method involves the balancing of species by checking their oxidation numbers. And correspondingly balancing all the atoms in the given reaction. If needed, water is also added to balance the equation.

Answer to Problem 53A

The balanced chemical equation is,

  $Cu\left(s\right)+4HN{O}_3\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+2N{O}_2\left(g\right)+2H_{2}O\left(l\right)$

Explanation of Solution

The given reaction,

  $Cu\left(s\right)+HN{O}_3\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$

The oxidation number of Cu changes to +2 from zero in $Cu{\left(N{O}_3\right)}_2$ by the loss of electrons, and oxidation. And the oxidation number of N in $HN{O}_3$ changes from +5 to +4 in $H_{2}O$ . The electron count can be balanced by multiplying the species involved in reduction with 2. The count of N is not equal on each side. So, it can be done by multiplying $HN{O}_3$ by 4 and 2 by $N{O}_2$ . The oxygen and hydrogen can be equalized by adding 2 to $H_{2}O$ . So, the equation becomes,

  $Cu\left(s\right)+4HN{O}_3\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+2N{O}_2\left(g\right)+2H_{2}O\left(l\right)$

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction, $P\left(s\right)+HN{O}_3\left(aq\right)+H_{2}O\to NO\left(g\right)+H_{3}P{O}_4\left(aq\right)$ is to be determined.

Concept Introduction: The oxidation number method involves the balancing of species by checking their oxidation numbers. And correspondingly balancing all the atoms in the given reaction. If needed, water is also added to balance the equation.

Answer to Problem 53A

The balanced chemical equation is,

  $3P\left(s\right)+5HN{O}_3\left(aq\right)+2H_{2}O\to 5NO\left(g\right)+3H_{3}P{O}_4\left(aq\right)$

Explanation of Solution

The given reaction,

  $P\left(s\right)+HN{O}_3\left(aq\right)+H_{2}O\to NO\left(g\right)+H_{3}P{O}_4\left(aq\right)$

The oxidation number of P changes from zero to +5 in $H_{3}P{O}_4$ . And the oxidation number of N changes from +5 in $HN{O}_3$ to +2 in NO. The electron count is not equal. So, it can be balanced by multiplying the oxidized species by 3 and reduced species by 5. So, the equation becomes,

  $3P\left(s\right)+5HN{O}_3\left(aq\right)+H_{2}O\to 5NO\left(g\right)+3H_{3}P{O}_4\left(aq\right)$

The oxygen count is not equal, so it can be balanced by multiplying $H_{2}O$ on the left-hand side by 2. So, the balanced equation is,

  $3P\left(s\right)+5HN{O}_3\left(aq\right)+2H_{2}O\to 5NO\left(g\right)+3H_{3}P{O}_4\left(aq\right)$

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation of the given reaction, $Bi{\left(OH\right)}_3\left(s\right)+N{a}_{2}Sn{O}_2\left(aq\right)\to Bi\left(s\right)+N{a}_{2}Sn{O}_3\left(aq\right)+H_{2}O\left(l\right)$ is to be determined.

Concept Introduction: The oxidation number method involves the balancing of species by checking their oxidation numbers. And correspondingly balancing all the atoms in the given reaction. If needed, water is also added to balance the equation.

Answer to Problem 53A

The balanced chemical equation is,

  $2Bi{\left(OH\right)}_3\left(s\right)+3N{a}_{2}Sn{O}_2\left(aq\right)\to \text{ 2}Bi\left(s\right)+3N{a}_{2}Sn{O}_3\left(aq\right)+3H_{2}O\left(l\right)$

Explanation of Solution

The given reaction,

  $Bi{\left(OH\right)}_3\left(s\right)+N{a}_{2}Sn{O}_2\left(aq\right)\to Bi\left(s\right)+N{a}_{2}Sn{O}_3\left(aq\right)+H_{2}O\left(l\right)$

The oxidation number of Bi changes from +3 of $Bi{\left(OH\right)}_3$ to zero. And the oxidation number of Sn changes from +2 to +4 by the loss of electrons in $N{a}_{2}Sn{O}_3$ . The electron count is not equal on both sides. It can be balanced by multiplying the species involved in reduction with 2 and the species involved in oxidation with 3. The equation therefore becomes,

  $2Bi{\left(OH\right)}_3\left(s\right)+3N{a}_{2}Sn{O}_2\left(aq\right)\to \text{ 2}Bi\left(s\right)+3N{a}_{2}Sn{O}_3\left(aq\right)+H_{2}O\left(l\right)$

Now, the count of oxygen can be balanced by multiplying the $H_{2}O$ by 3. So, the balanced equation is,

  $2Bi{\left(OH\right)}_3\left(s\right)+3N{a}_{2}Sn{O}_2\left(aq\right)\to \text{ 2}Bi\left(s\right)+3N{a}_{2}Sn{O}_3\left(aq\right)+3H_{2}O\left(l\right)$