Chapter 3, Problem 141QRT

(a)

Interpretation Introduction

Interpretation:

From the given trials of reactants, a trial that contains Stoichiometric amounts of the reactants has to be determined.

Explanation of Solution

Given,

Trial	A	B	C	D	E
$\text{0}{\text{.10 M X}}^{\text{n+}}$	$\text{7 mL}$	$\text{6 mL}$	$\text{5 mL}$	$\text{4 mL}$	$\text{3 mL}$
$\text{0}{\text{.20 M Y}}^{\text{m-}}$	$\text{3 mL}$	$\text{4 mL}$	$\text{5 mL}$	$\text{6 mL}$	$\text{7 mL}$
Excess ${\text{X}}^{\text{n+}}$ present	Yes	Yes	Yes	No	No
Excess ${\text{Y}}^{\text{m-}}$ present	No	No	No	No	Yes

By observing the given trails, Trial D has no excess reactants present in stoichiometric amounts of reactants.

Assume the volumes are measured with at least graduated cylinders, giving them one decimal place.

(b)

Interpretation Introduction

Interpretation:

Number of moles of ${\text{X}}^{\text{n+}}$ reacted in the trials has to be calculated.

Concept introduction:

Number of moles is calculated as follows,

  $\text{Number}\text{of}\text{moles}\text{=}\text{Volume}\text{of}\text{solution}\text{in}\text{litre}\text{×}\text{Concentration}$

Explanation of Solution

Given,

Trial	A	B	C	D	E
$\text{0}{\text{.10 M X}}^{\text{n+}}$	$\text{7 mL}$	$\text{6 mL}$	$\text{5 mL}$	$\text{4 mL}$	$\text{3 mL}$
$\text{0}{\text{.20 M Y}}^{\text{m-}}$	$\text{3 mL}$	$\text{4 mL}$	$\text{5 mL}$	$\text{6 mL}$	$\text{7 mL}$
Excess ${\text{X}}^{\text{n+}}$ present	Yes	Yes	Yes	No	No
Excess ${\text{Y}}^{\text{m-}}$ present	No	No	No	No	Yes

Number of moles of ${\text{X}}^{\text{n+}}$ reacted in the trials is calculated as follows,

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\text{Volume}\text{of}\text{solution}\text{in}\text{litre}\text{×}\text{Concentration}\\ \text{Number}\text{of}\text{moles}\text{=}\text{4}{\text{.0×10}}^{\text{-3}}\text{L×}\frac{\text{0}\text{.10}\text{mol}{\text{X}}^{\text{n+}}}{\text{1}\text{L}}\\ \text{Number}\text{of}\text{moles}\text{=}\text{4}{\text{.0×10}}^{\text{-4}}\text{mol}{\text{X}}^{\text{n+}}\end{array}$

Number of moles is calculated.

(c)

Interpretation Introduction

Interpretation:

Number of moles of ${\text{Y}}^{\text{m-}}$ reacted in the trials has to be calculated.

Concept introduction:

  Refer to part (b)

Explanation of Solution

Given,

Trial	A	B	C	D	E
$\text{0}{\text{.10 M X}}^{\text{n+}}$	$\text{7 mL}$	$\text{6 mL}$	$\text{5 mL}$	$\text{4 mL}$	$\text{3 mL}$
$\text{0}{\text{.20 M Y}}^{\text{m-}}$	$\text{3 mL}$	$\text{4 mL}$	$\text{5 mL}$	$\text{6 mL}$	$\text{7 mL}$
Excess ${\text{X}}^{\text{n+}}$ present	Yes	Yes	Yes	No	No
Excess ${\text{Y}}^{\text{m-}}$ present	No	No	No	No	Yes

Number of moles of ${\text{Y}}^{\text{m-}}$ reacted in the trials is calculated as follows,

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\text{Volume}\text{of}\text{solution}\text{in}\text{litre}\text{×}\text{Concentration}\\ \text{Number}\text{of}\text{moles}\text{=}\text{6}{\text{.0×10}}^{\text{-3}}\text{L×}\frac{\text{0}\text{.20}\text{mol}{\text{Y}}^{\text{m-}}}{\text{1}\text{L}}\\ \text{Number}\text{of}\text{moles}\text{=}\text{1}{\text{.2×10}}^{\text{-3}}\text{mol}{\text{Y}}^{\text{m-}}\end{array}$

Number of moles is calculated.

(d)

Interpretation Introduction

Interpretation:

The whole number ratio of ${\text{X}}^{\text{n+}}$ to ${\text{Y}}^{\text{m-}}$ reacted has to be calculated.

Concept introduction:

  Refer to part (b)

Explanation of Solution

Given,

Trial	A	B	C	D	E
$\text{0}{\text{.10 M X}}^{\text{n+}}$	$\text{7 mL}$	$\text{6 mL}$	$\text{5 mL}$	$\text{4 mL}$	$\text{3 mL}$
$\text{0}{\text{.20 M Y}}^{\text{m-}}$	$\text{3 mL}$	$\text{4 mL}$	$\text{5 mL}$	$\text{6 mL}$	$\text{7 mL}$
Excess ${\text{X}}^{\text{n+}}$ present	Yes	Yes	Yes	No	No
Excess ${\text{Y}}^{\text{m-}}$ present	No	No	No	No	Yes

Number of moles of ${\text{X}}^{\text{n+}}$ reacted in the trials is calculated as follows,

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\text{Volume}\text{of}\text{solution}\text{in}\text{litre}\text{×}\text{Concentration}\\ \text{Number}\text{of}\text{moles}\text{=}\text{4}{\text{.0×10}}^{\text{-3}}\text{L×}\frac{\text{0}\text{.10}\text{mol}{\text{X}}^{\text{n+}}}{\text{1}\text{L}}\\ \text{Number}\text{of}\text{moles}\text{=}\text{4}{\text{.0×10}}^{\text{-4}}\text{mol}{\text{X}}^{\text{n+}}\end{array}$

Number of moles of ${\text{Y}}^{\text{m-}}$ reacted in the trials is calculated as follows,

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\text{Volume}\text{of}\text{solution}\text{in}\text{litre}\text{×}\text{Concentration}\\ \text{Number}\text{of}\text{moles}\text{=}\text{6}{\text{.0×10}}^{\text{-3}}\text{L×}\frac{\text{0}\text{.20}\text{mol}{\text{Y}}^{\text{m-}}}{\text{1}\text{L}}\\ \text{Number}\text{of}\text{moles}\text{=}\text{1}{\text{.2×10}}^{\text{-3}}\text{mol}{\text{Y}}^{\text{m-}}\end{array}$

The whole number ratio of ${\text{X}}^{\text{n+}}$ to ${\text{Y}}^{\text{m-}}$ is given below,

  The determination of mole ratio is = $\text{4}{\text{.0×10}}^{\text{-4}}\text{mol}{\text{X}}^{\text{n+}}:\text{1}{\text{.2×10}}^{\text{-3}}\text{mol}{\text{Y}}^{\text{m-}}$

  Dividing by 4,

  The determination of mole ratio is = ${\text{1×10}}^{\text{-4}}\text{mol}{\text{X}}^{\text{n+}}:3{\text{×10}}^{\text{-4}}\text{mol}{\text{Y}}^{\text{m-}}$

The whole number ratio of ${\text{X}}^{\text{n+}}$ to ${\text{Y}}^{\text{m-}}$ is ${\text{1×10}}^{\text{-4}}\text{mol}{\text{X}}^{\text{n+}}:3{\text{×10}}^{\text{-4}}\text{mol}{\text{Y}}^{\text{m-}}$.

(d)

Interpretation Introduction

Interpretation:

The chemical formula of the product has to be determined.

Explanation of Solution

The whole number ratio of ${\text{X}}^{\text{n+}}$ to ${\text{Y}}^{\text{m-}}$ is ${\text{1×10}}^{\text{-4}}\text{mol}{\text{X}}^{\text{n+}}:3{\text{×10}}^{\text{-4}}\text{mol}{\text{Y}}^{\text{m-}}$.

According to the ratio calculation, x is ${\text{1×10}}^{\text{-4}}\text{mol}{\text{X}}^{\text{n+}}$ mole and y is $3{\text{×10}}^{\text{-4}}\text{mol}{\text{Y}}^{\text{m-}}$. Therefore the chemical formula of the product is ${\text{XY}}_{\text{3}}$.