Chapter 9, Problem 16A

(a)

Interpretation Introduction

Interpretation:

The number of moles of second reactant that would be required to react completely with exactly 25.0 g of the first reactant needs to be determined.

  $\text{Mg}\left(\text{s}\right){\text{ + CuCl}}_{\text{2}}\left(\text{aq}\right)\to {\text{ MgCl}}_{\text{2}}\left(\text{aq}\right)\text{ + Cu}\left(\text{s}\right)$

Concept Introduction:

From the balanced equation, moles of second reactant needed to react with 25g of first reactant are calculated.

  $\text{Number of moles =}\frac{\text{Mass}}{\text{Molar mass}}$

Answer to Problem 16A

Moles of CuCl₂required = 1.041moles

Explanation of Solution

  $\text{Mg}\left(\text{s}\right){\text{ + CuCl}}_{\text{2}}\left(\text{aq}\right)\to {\text{ MgCl}}_{\text{2}}\left(\text{aq}\right)\text{ + Cu}\left(\text{s}\right)$

Mass of first reactant = 25.0 g

Molar mass of Mg = 24.0 g/mol

  $\text{Number of moles =}\frac{25}{24}=1.041mol$

Since the stoichiometric coefficients of Mg and CuCl₂ are the same, 1 mole of Mg will require 1 mole of CuCl_2.

Hence, moles of CuCl₂required = 1.041moles

(b)

Interpretation Introduction

Interpretation:

The number of moles of the second reactant that would be required to react completely with exactly 25.0 g of the first reactant needs to be determined.

  ${\text{ 2AgNO}}_{\text{3}}{\text{+ NiCl}}_{\text{2}}\to \text{ 2AgCl + Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

Concept Introduction:

From the balanced equation, moles of the second reactant needed to react with 25g of the first reactant are calculated.

  $\text{Number of moles =}\frac{Mass}{Molar\text{ mass}}$

Answer to Problem 16A

0.147 moles of AgNO₃ will require= 0.735moles

Explanation of Solution

  ${\text{ 2AgNO}}_{\text{3}}{\text{+ NiCl}}_{\text{2}}\to \text{ 2AgCl + Ni}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

  $\begin{array}{l}{\text{Number of moles KHCO}}_{\text{3}}\text{=}\frac{\text{Mass}}{\text{Molar mass}}\text{=}\frac{\text{25}}{\text{168}\text{.87}}\text{=0}\text{.147moles}\\ {\text{2 moles of AgNO}}_{\text{3}}{\text{reacts with 1 mole of NiCl}}_{\text{2}}\text{,}\\ \text{0}{\text{.147 moles of AgNO}}_{\text{3}}\text{will require=}\frac{\text{1}}{\text{2}}\text{×0}\text{.147=0}\text{.0735moles}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of the second reactant that would be required to react completely with exactly 25.0 g of the first reactant needs to be determined.

  ${\text{ NaHSO}}_{\text{3}}\text{ + NaOH }\to {\text{ Na}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}$

Concept Introduction:

From the balanced equation, moles of the second reactant needed to react with 25g of the first reactant is calculated.

  $\text{Number of moles =}\frac{\text{Mass}}{\text{Molar mass}}$

Answer to Problem 16A

0.24 moles of NaHSO₃require 0.24 moles of NaOH.

Explanation of Solution

  ${\text{ NaHSO}}_{\text{3}}\text{ + NaOH }\to {\text{ Na}}_{\text{2}}{\text{SO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}$

  ${\text{Number of moles NaHSO}}_3\text{=}\frac{Mass}{Molar\text{ mass}}=\frac{25}{104}=0.24moles$

1 mole of NaHSO3 requires 1 mole of NaOH.

Hence, 0.24 moles of NaHSO3 require 0.24 moles of NaOH.

(d)

Interpretation Introduction

Interpretation:

The number of moles of the second reactant that would be required to react completely with exactly 25.0 g of the first reactant needs to be determined.

  ${\text{KHCO}}_{\text{3}}\text{ + HCl }\to {\text{ KCl + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

Concept Introduction:

From the balanced equation, moles of the second reactant needed to react with 25g of the first reactant is calculated.

  $\text{Number of moles =}\frac{Mass}{Molar\text{ mass}}$

Answer to Problem 16A

0.25 mole requires 0.25 mole of HCl.

Explanation of Solution

  ${\text{KHCO}}_{\text{3}}\text{ + HCl }\to {\text{ KCl + H}}_{\text{2}}{\text{O + CO}}_{\text{2}}$

  ${\text{Number of moles KHCO}}_3\text{=}\frac{Mass}{Molar\text{ mass}}=\frac{25}{100}=0.25moles$

1 mole of KHCO₃ requires 1 mole of HCl.

Hence, 0.25 mole requires 0.25 mole of HCl.