Chapter 4, Problem 3MCP

(a)

Interpretation Introduction

Interpretation:

Whether the reaction that takes place by mixing of sodium phosphate and copper $\text{(II)}$ sulphate is an acid-base reaction, or redox or precipitation reaction has to be given.

Concept Introduction:

Precipitation reaction:

Precipitation reaction is a reaction in which any chemical change in a solution results in giving one or more insoluble products.  The insoluble product is called as precipitate.

Acid-Base reaction:

It involves the transfer of hydrogen ion, from one reactant (the acid) to another (the base).

Oxidation-Reduction reactions:

The transfer of negative charge (one or more electrons) from one reactant to other reactant results in oxidation-reduction reactions.  When a subtance/species lose electrons during a reaction, then it is a oxidation reaction.  When a subtance/species gain electrons during a reaction, then it is a reduction reaction.

Oxidation reaction and reduction reaction are complementary reactions and are called as redox reactions.

Explanation of Solution

Given:

A reaction is carried out by mixing sodium phosphate and copper $\text{(II)}$ sulphate in water.

On mixing sodium phosphate and copper $\text{(II)}$ sulphate in water in water, sodium sulphate and copper phosphate are formed.  Copper phosphate is obtained as insoluble precipitate and hence this reaction is a precipitation reaction.

The chemical reaction equation for the above reaction is,

  ${\text{3CuSO}}_{\text{4(aq)}}{\text{+ 2Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 3Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$

(b)

Interpretation Introduction

Interpretation:

The reaction of sodium phosphate and copper $\text{(II)}$ sulphate has to be classified as decomposition (D), combination (C), single-replacement (SR) or double-replacement (DR).

Concept Introduction:

Combination reaction (C) is a reaction in which two or more elements or compounds join together producing a product of different composition.  The general form of a combination reaction is $\text{A + B }\stackrel{}{\to }\text{ AB}$.

Decomposition reaction (D) is a reaction in which a single reactant undergoes decomposition and gives two or more products.  The general form of a decomposition reaction is $\text{AB }\stackrel{}{\to }\text{ A + B}$.

Single-replacement reaction (SR) is a reaction in which one atom replaces another in the compound and produces a new compound.  The general form of a single-replacement reaction is $\text{A + BC }\stackrel{}{\to }\text{ AC + B}$.

Double-replacement reaction (DR) is a reaction in which two compounds undergoes “change of partners”.  That is, two compounds react by exchanging atoms to produce two new compounds.  The general form of a double-replacement reaction is $\text{AB + CD }\stackrel{}{\to }\text{ AD + CB}$.

Explanation of Solution

Given:

A reaction is carried out by mixing sodium phosphate and copper $\text{(II)}$ sulphate in water.

The chemical reaction equation for the above reaction is,

  ${\text{3CuSO}}_{\text{4(aq)}}{\text{+ 2Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 3Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$

In the above reaction, both the reactants exchange their partners with each other.  ${\text{CuSO}}_{\text{4}}$ gives ${\text{SO}}_{\text{4}}{}^{2-}$ and accepts ${\text{PO}}_{\text{4}}{}^{3-}$ from ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$.  Similarly, ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ gives ${\text{PO}}_{\text{4}}{}^{3-}$ and accepts ${\text{SO}}_{\text{4}}{}^{2-}$ from ${\text{CuSO}}_{\text{4}}$.

Hence, this reaction is a Double-replacement reaction (DR).

(c)

Interpretation Introduction

Interpretation:

The net ionic equation, the balanced equation and the ionic equation for the given reaction has to be written.

Concept Introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change.  The ions that do not change during the chemical reaction are called spectator ions.  These spectator ions are not shown in net ionic equation.

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

Explanation of Solution

Given:

A reaction is carried out by mixing sodium phosphate and copper $\text{(II)}$ sulphate in water.

Balanced equation:

The balanced chemical equation for the given reaction is written as,

  ${\text{3CuSO}}_{\text{4(aq)}}{\text{+ 2Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 3Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$

The ionic equation for the above reaction is written as,

  ${\text{3Cu}}_{(\text{aq)}}^{2+}{\text{+3SO}}_{4\text{(aq)}}^{2-}{\text{ +6Na}}_{\text{(aq)}}^{+}{\text{+2PO}}_{4\text{(aq)}}^{3-}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 6Na}}_{\text{(aq)}}^{+}+{\text{3SO}}_{4\text{(aq)}}^{2-}$

The net ionic equation is obtained by cancelling the spectator ions in the ionic equation.

  ${\text{3Cu}}_{(\text{aq)}}^{2+}\text{+}\cancel{{\text{3SO}}_{4\text{(aq)}}^{2-}}\text{ +}\cancel{{\text{6Na}}_{\text{(aq)}}^{+}}{\text{+2PO}}_{4\text{(aq)}}^{3-}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}\text{+ }\cancel{{\text{6Na}}_{\text{(aq)}}^{+}}+\cancel{{\text{3SO}}_{4\text{(aq)}}^{2-}}$

Therefore, the net ionic equation of the given reaction is written as,

  ${\text{3Cu}}_{(\text{aq)}}^{2+}{\text{+2PO}}_{4\text{(aq)}}^{3-}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}$

(d)

Interpretation Introduction

Interpretation:

The number of moles of sodium phosphate that is represented by $\text{6}\text{.00 g}$ of sodium phosphate has to be calculated.

Concept Introduction:

Moles:

Mole of the substance is found by dividing the mass of the substance by its molar mass.

  $\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

Explanation of Solution

Given:

A reaction is carried out by mixing sodium phosphate and copper $\text{(II)}$ sulphate in water.

The mass of sodium phosphate is $\text{6}\text{.00 g}$

Calculation of number of moles:

The balanced chemical equation for the given reaction is written as,

  ${\text{3CuSO}}_{\text{4(aq)}}{\text{+ 2Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 3Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$

The molar mass of sodium phosphate is $\text{163}\text{.94 g/mol}$

Determine the moles of sodium phosphate present in given amount as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{6}\text{.00 g}}{\text{163}\text{.94 g/mol}}\\ =0.0366\text{ mol}\end{array}$

The number of moles of sodium phosphate present is $0.0366\text{ moles}$

(e)

Interpretation Introduction

Interpretation:

The number of moles of copper $\text{(II)}$ sulfate that must be present for the complete reaction to occur has to be given.

Explanation of Solution

Given:

A reaction is carried out by mixing sodium phosphate and copper $\text{(II)}$ sulphate in water.

The balanced chemical equation for the given reaction is written as,

  ${\text{3CuSO}}_{\text{4(aq)}}{\text{+ 2Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 3Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$

From the stiochiometry of the reaction equation, it is known that three moles of copper $\text{(II)}$ sulfate reacts with two moles of sodium phosphate.

Hence, the number of moles of copper $\text{(II)}$ sulfate that is required for the complete reaction to occur is ${\text{3 mol CuSO}}_4$.

(f)

Interpretation Introduction

Interpretation:

The number of moles of products produced in the given reaction has to be given.

Explanation of Solution

Given:

A reaction is carried out by mixing sodium phosphate and copper $\text{(II)}$ sulphate in water.

The balanced chemical equation for the given reaction is written as,

  ${\text{3CuSO}}_{\text{4(aq)}}{\text{+ 2Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 3Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$

From the stiochiometry of the reaction equation, it is known that three moles of copper $\text{(II)}$ sulfate reacts with two moles of sodium phosphate.

Hence, the number of moles of copper $\text{(II)}$ sulfate that is required for the complete reaction to occur is ${\text{3 mol CuSO}}_4$.

From the stiochiometry of the reaction equation, it is known that three moles of copper $\text{(II)}$ sulfate reacts with two moles of sodium phosphate and produces one mole of copper $\text{(II)}$ phosphate and three moles of sodium sulfate as product.

Hence, the number of moles of products produced in the given reaction are,

  $\begin{array}{l}{\text{1 mol Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\\ {\text{3 mol Na}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

(g)

Interpretation Introduction

Interpretation:

The amount (in $\text{g}$) of products produced in the given reaction has to be calculated.

Concept Introduction:

Mass:

Mass of the compound is calculated by mole of the compound multiplied with molar mass of the compound.

  $\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Explanation of Solution

Given:

A reaction is carried out by mixing sodium phosphate and copper $\text{(II)}$ sulphate in water.

The balanced chemical equation for the given reaction is written as,

  ${\text{3CuSO}}_{\text{4(aq)}}{\text{+ 2Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}{\text{+ 3Na}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$

From the stiochiometry of the reaction equation, it is known that three moles of copper $\text{(II)}$ sulfate reacts with two moles of sodium phosphate and produces one mole of copper $\text{(II)}$ phosphate and three moles of sodium sulfate as product.

Hence, the number of moles of copper $\text{(II)}$ phosphate and sodium sulfate produced in the given reaction is,

  $\begin{array}{l}{\text{1 mol Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\\ {\text{3 mol Na}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar mass of copper $\text{(II)}$ phosphate is $\text{380}\text{.58 g/mol}$.

Determine the mass of copper $\text{(II)}$ phosphate formed as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{ Molar}\text{mass}\text{×}\text{mole}\\ \text{= 380}\text{.58 g/mol}\times {\text{1 mol Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\\ =\text{380}\text{.58 g}\\ =\text{ 380}{\text{.58 g Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\end{array}$

The molar mass of sodium sulfate is $\text{142}\text{.04 g/mol}$.

Determine the mass of sodium sulfate formed as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{ Molar}\text{mass}\text{×}\text{mole}\\ \text{= 142}\text{.04 g/mol}\times {\text{3 mol Na}}_{\text{2}}{\text{SO}}_{\text{4}}\\ =426.12\text{ g}\\ =\text{ 426}{\text{.1 g Na}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Therefore, the mass of products produced is $\text{380}{\text{.58 g Cu}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ and $\text{426}{\text{.1 g Na}}_{\text{2}}{\text{SO}}_{\text{4}}$