Chapter 9, Problem 20CR

Solid calcium carbide $\left({\text{CaC}}_2\right)$reacts with liquid water to produce acetylene gas $\left({\text{C}}_2{\text{H}}_2\right)$and aqueous calcium hydroxide.

l type='a'>

Write the balanced equation for the reaction that is (occurring. including all phases.

If a 100.0-g sample of calcium carbide $\left({\text{CaC}}_2\right)$is initially reacted with 50.0 g of water, which reactant is limiting?

Prove that mass is conserved for the reactant amounts used in pan b.

Interpretation Introduction

(a)

Interpretation:

The balanced chemical reaction for the solid calcium carbide with liquid water should be determined.

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 20CR

The balance equation for the solid calcium carbide with liquid water is as follows:

${\text{CaC}}_{\text{2}}(s)+2{\text{H}}_{\text{2}}\text{O}(l)\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}(g)+{\text{ Ca(OH)}}_2(aq)$.

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 equation for the solid calcium carbide with liquid water is as follows:

${\text{CaC}}_{\text{2}}(s)+{\text{2H}}_{\text{2}}\text{O}(l)\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}(g)+{\text{ Ca(OH)}}_{\text{2}}(aq)$.

Interpretation Introduction

(b)

Interpretation:

The limiting reagent should be determined.

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 20CR

${\text{H}}_{\text{2}}\text{O}$ is a limiting agent.

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 equation for the solid calcium carbide with liquid water is as follows:

${\mathrm{CaC}}_2(s)+2{\text{H}}_2\text{O}(l)\to {\text{C}}_2{\text{H}}_2(g)+\text{C}\text{a}{(\text{O}\text{H})}_2(aq)$

Given:

Amount of ${\text{CaC}}_{\text{2}}$ = 100.0 g

Amount of ${\text{H}}_{\text{2}}\text{O}$ = 50.0 g

Calculation:

Number of moles of $\text{C}\text{a}{\text{C}}_2$ and ${\text{H}}_2\text{O}$ calculated as follows:

$\begin{array}{l}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}=\frac{\text{m}\text{a}\text{s}\text{s}\text{i}\text{n}\text{g}}{\text{m}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}}=\frac{100.0\text{g}}{64.099\text{g}/\text{m}\text{o}\text{l}}=1.56\text{m}\text{o}\text{l}\text{e}\text{s}\text{C}\text{a}{\text{C}}_2\\ \text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}=\frac{\text{m}\text{a}\text{s}\text{s}\text{i}\text{n}\text{g}}{\text{m}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}}=\frac{50.0\text{g}}{18.02\text{g}/\text{m}\text{o}\text{l}}=2.77\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}\end{array}$

Amount of other reactant is calculated as follows:

$\text{2}{\text{.77 moles H}}_{\text{2}}\text{O×}\frac{\text{ 1}{\text{.00 moles CaC}}_{\text{2}}}{\text{2}{\text{.00 moles H}}_{\text{2}}\text{O}}\text{ = 1}{\text{.385 moles CaC}}_{\text{2}}$

Here, water is a limiting agent and calcium carbide present in excess.

Interpretation Introduction

(c)

Interpretation:

To prove that mass in conserved for reactant

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 20CR

Both sides amount approximately equal thus that mass in conserved for reactant.

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 equation for the solid calcium carbide with liquid water is as follows:

${\mathrm{CaC}}_2(s)+2{\text{H}}_2\text{O}(l)\to {\text{C}}_2{\text{H}}_2(g)+\text{C}\text{a}{(\text{O}\text{H})}_2(aq)$

Given:

Amount of ${\text{CaC}}_{\text{2}}$ = 100.0 g

Amount of ${\text{H}}_{\text{2}}\text{O}$ = 50.0 g

Calculation:

Number of moles of ${\text{CaC}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ calculated as follows:

$\begin{array}{l}\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}=\frac{\text{m}\text{a}\text{s}\text{s}\text{i}\text{n}\text{g}}{\text{m}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}}=\frac{100.0\text{g}}{64.099\text{g}/\text{m}\text{o}\text{l}}=1.56\text{m}\text{o}\text{l}\text{e}\text{s}\text{C}\text{a}{\text{C}}_2\\ \text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}=\frac{\text{m}\text{a}\text{s}\text{s}\text{i}\text{n}\text{g}}{\text{m}\text{o}\text{l}\text{a}\text{r}\text{m}\text{a}\text{s}\text{s}}=\frac{50.0\text{g}}{18.02\text{g}/\text{m}\text{o}\text{l}}=2.77\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}\end{array}$

Amount of other reactant which left after reaction is calculated as follows:

$\begin{array}{l}2.77\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}\times \frac{1.00\text{m}\text{o}\text{l}\text{e}\text{s}\text{C}\text{a}{\text{C}}_2}{2.00\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}}\times 1.385\text{m}\text{o}\text{l}\text{e}\text{s}\text{C}\text{a}{\text{C}}_2\times 64.099\text{g}/\text{m}\text{o}\text{l}=88.77\text{g}\text{C}\text{a}{\text{C}}_2\\ 100.0\text{g}\text{C}\text{a}{\text{C}}_2-88.77\text{g}\text{C}\text{a}{\text{C}}_2=11.23\text{g}\text{C}\text{a}{\text{C}}_2\end{array}$

Amount of product in gram calculated as follows:

$\begin{array}{l}2.77\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}\times \frac{\cancel{1.00\text{m}\text{o}\text{l}\text{e}{\text{C}}_2{\text{H}}_2}}{2.00\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}}}\times \frac{26.04\text{g}{\text{C}}_2{\text{H}}_2}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}{\text{C}}_2{\text{H}}_2}}=26.0654\text{g}{\text{C}}_2{\text{H}}_2\\ 2.77\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}\times \frac{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{C}\text{a}{(\text{O}\text{H})}_2}}{2.00\cancel{\text{m}\text{o}\text{l}\text{e}\text{s}{\text{H}}_2\text{O}}}\times \frac{74.093\text{g}\text{C}\text{a}{(\text{O}\text{H})}_2}{\cancel{1.00\text{m}\text{o}\text{l}\text{e}\text{C}\text{a}{(\text{O}\text{H})}_2}}=102.62\text{g}\text{C}\text{a}{(}_{\text{O}}\end{array}$

Now;

Amount of reaming reactant + total product amount = amount of used reactants

$\begin{array}{l}\text{11}{\text{.23 g CaC}}_{\text{2}}\text{+26}{\text{.0654 g C}}_{\text{2}}{\text{H}}_{\text{2}}\text{+102}{\text{.62 g Ca(OH)}}_{\text{2}}\text{=88}{\text{.77g CaC}}_{\text{2}}\text{+50}{\text{.0 g H}}_{\text{2}}\text{O}\\ \text{139}\text{.92g=138}\text{.77g}\end{array}$

Both sides amount approximately equal thus that mass in conserved for reactant.