Chapter 9, Problem 18CR

Interpretation Introduction

(a)

Interpretation:

The amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

$\begin{array}{l}2\text{ PbS }+{\text{ 3 O}}_2\to 2\text{PbO }+2{\text{SO}}_{\text{2}}\\ \text{ Reactants Products}\end{array}$

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 18CR

There are $5.3{\text{ g SiCl}}_{\text{4}}$ and $\text{0}\text{.375 g C}$ formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

$\text{SiC}(s)+{\text{2Cl}}_{\text{2}}(g)\to {\text{SiCl}}_{\text{4}}(l)+\text{C}(s)$

Given:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of $\text{S}\text{i}\text{C}$ calculated as follows:

$\text{Number of moles}\text{=}\frac{\text{m}\text{a}\text{s}\text{s}\text{i}\text{n}\text{g}}{\text{m}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}}=\frac{1.25\text{g}}{40.11\text{g}/\text{m}\text{o}\text{l}}=0.0312\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\text{i}\text{C}$

Amount of product in gram calculated as follows:

$0.0312\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\text{i}\text{C}}\times \frac{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{S}\text{i}\text{C}{\text{l}}_4}}{1.00\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\text{i}\text{C}}}\times \frac{169.9\text{g}\text{S}\text{i}\text{C}{\text{l}}_4}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{S}\text{i}\text{C}{\text{l}}_4}}=5.3{\text{ g SiCl}}_{\text{4}}$

$0.0312\cancel{\text{moles SiC}}\text{×}\frac{\cancel{\text{1}\text{.00 mole C}}}{\text{1}\text{.00 }\cancel{\text{moles SiC}}}\text{×}\frac{\text{12}\text{.01 g C}}{\cancel{\text{1}\text{.00 mole C}}}=0.375\text{ g C}$.

Interpretation Introduction

(b)

Interpretation:

The amount of product in the given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

$\begin{array}{l}2\text{ PbS }+{\text{ 3 O}}_2\to 2\text{PbO }+2{\text{SO}}_{\text{2}}\\ \text{ Reactants Products}\end{array}$

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 18CR

There is $2.00\text{g}\text{L}\text{i}\text{O}\text{H}$ formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

${\text{Li}}_{\text{2}}\text{O}(s)+{\text{H}}_{\text{2}}\text{O(}\text{l})\to 2\text{LiOH}(aq)$

Given:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of ${\text{Li}}_{\text{2}}\text{O}$ calculated as follows:

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{m}\text{a}\text{s}\text{s}\text{i}\text{n}\text{g}}{\text{m}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}}\\ =\frac{1.25\text{g}}{29.88\text{g}/\text{m}\text{o}\text{l}}\\ =0.0418\text{m}\text{o}\text{l}\text{e}\text{s}\text{L}{\text{i}}_2\text{O}\end{array}$

Amount of product in gram calculated as follows:

$0.0418\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}\text{L}{\text{i}}_2\text{O}}\times \frac{\cancel{2.00\text{m}\text{o}\text{l}\text{e}\text{L}\text{i}\text{O}\text{H}}}{1.00\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}\text{L}{\text{i}}_2\text{O}}}\times \frac{23.95\text{g}\text{L}\text{i}\text{O}\text{H}}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{L}\text{i}\text{O}\text{H}}}=2.00\text{g}\text{L}\text{i}\text{O}\text{H}$.

Interpretation Introduction

(c)

Interpretation:

To calculate the amount of product in given reaction

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

$\begin{array}{l}2\text{ PbS }+{\text{ 3 O}}_2\to 2\text{PbO }+2{\text{SO}}_{\text{2}}\\ \text{ Reactants Products}\end{array}$

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 18CR

There are $\text{2}\text{.56 g NaOH}$ and $\text{0}{\text{.512 g O}}_{\text{2}}$ formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

$2{\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}(s)+{\text{2H}}_{\text{2}}\text{O}(l)\to 4\text{NaOH}(aq)+{\text{O}}_2(g)$

Given:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of ${\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}$ calculated as follows:

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{\text{1}\text{.25 g}}{\text{77}\text{.98 g/mol}}\text{=0}{\text{.016 molesNa}}_{\text{2}}{\text{O}}_{\text{2}}$

Amount of product in gram calculated as follows:

$\begin{array}{l}0.016\cancel{{\text{molesNa}}_{\text{2}}{\text{O}}_{\text{2}}}\text{×}\frac{\cancel{\text{4}\text{.00 mole NaOH}}}{\text{1}\text{.00 }\cancel{{\text{molesNa}}_{\text{2}}{\text{O}}_{\text{2}}}}\text{×}\frac{\text{39}\text{.997 g NaOH}}{\cancel{\text{1}\text{.00 mole NaOH}}}=2.56\text{ g NaOH}\\ 0.016\cancel{{\text{molesNa}}_{\text{2}}{\text{O}}_{\text{2}}}\text{×}\frac{\cancel{\text{1}{\text{.00 mole O}}_{\text{2}}}}{\text{1}\text{.00 }\cancel{{\text{molesNa}}_{\text{2}}{\text{O}}_{\text{2}}}}\text{×}\frac{\text{32}{\text{.00 g O}}_{\text{2}}}{\cancel{\text{1}{\text{.00 mole O}}_{\text{2}}}}=0.5{\text{12 g O}}_{\text{2}}\end{array}$.

Interpretation Introduction

(d)

Interpretation:

The amount of product in given reaction should be calculated.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

$\begin{array}{l}2\text{ PbS }+{\text{ 3 O}}_2\to 2\text{PbO }+2{\text{SO}}_{\text{2}}\\ \text{ Reactants Products}\end{array}$

Mass of any substance can be calculated as follows:

$\text{Mass in gram = Number of moles}\times \text{Molar mass}$

Number of moles can be calculated as follows;

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}$.

Answer to Problem 18CR

There are $\text{0}\text{.984 g Sn}$ and $\text{0}{\text{.299 g H}}_{\text{2}}\text{O}$ formed in the reaction.

Explanation of Solution

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.

If any reactant left after competitions of reaction, thus it is said to excess reactant.

The balance chemical equation is as follows:

${\text{SnO}}_{\text{2}}(s)+{\text{2H}}_{\text{2}}(g)\to \text{Sn}(s)+{\text{2H}}_{\text{2}}\text{O}(l)$

Given Information:

Amount of first reactant = 12.5 g

Calculation:

Number of moles of $\text{S}\text{n}{\text{O}}_2$ calculated as follows:

$\text{Number of moles}\text{=}\frac{\text{mass in g}}{\text{molar}\text{mass}}=\frac{1.25\text{g}}{150.71\text{g}/\text{m}\text{o}\text{l}}=0.00829{\text{ moles SnO}}_{\text{2}}$

Amount of product in gram calculated as follows:

$\begin{array}{l}0.00829\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\text{n}{\text{O}}_2\times \frac{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{S}\text{n}}}{1.00\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\text{n}{\text{O}}_2}}\times \frac{118.71\text{g}\text{S}\text{n}}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{S}\text{n}}}=0.984\text{g}\text{S}\text{n}\\ 0.00829\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\text{n}{\text{O}}_2\times \frac{\cancel{2.00\text{m}\text{o}\text{l}\text{e}{\text{H}}_2\text{O}}}{1.00\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}\text{S}\text{n}{\text{O}}_2}}\times \frac{18.02\text{g}{\text{H}}_2\text{O}}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}{\text{H}}_2\text{O}}}=0.299\text{g}{\text{H}}_2\text{O}\end{array}$.