End of chapter exercises | Quantitative aspects of chemical change

8.4 Chapter summary

9.1 Introduction

Textbook Exercise 8.9

A reagent that is completely used up in a chemical reaction.

Limiting reagent

The simplest formula of a compound.

Empirical formula

The amount of product that is calculated for a reaction using stoichiometric methods.

Theoretical yield

A technique for determining the concentration of an unknown solution.

Titration

What is the volume of $\text{3}$ $\text{mol}$ of $\text{N}_{2}$ gas at STP?

$\text{67,2}$ $\text{dm$^{3}$}$
$\text{22,4}$ $\text{dm$^{3}$}$
$\text{33,6}$ $\text{dm$^{3}$}$
$\text{63,2}$ $\text{dm$^{3}$}$

$\text{67,2}$ $\text{dm$^{3}$}$

Given the following reaction:

\[3\text{NO}_{2}\text{(g)} + \text{H}_{2}\text{O(g)} \rightarrow 2\text{HNO}_{3} \text{(g)} + \text{NO(g)}\]

If $\text{2,7}$ $\text{dm$^{3}$}$ of $\text{NO}_{2}$ is used, what volume of $\text{HNO}_{3}$ is produced at STP?

$\text{4,1}$ $\text{dm$^{3}$}$
$\text{2,7}$ $\text{dm$^{3}$}$
$\text{1,8}$ $\text{dm$^{3}$}$
$\text{3,4}$ $\text{dm$^{3}$}$

$\text{1,8}$ $\text{dm$^{3}$}$

\begin{align*} V_{A} & = \frac{a}{b}V_{B} \\ V_{\text{HNO}_{3}} & = \frac{2}{3}V_{\text{NO}_{2}} \\ & = \frac{2}{3}(\text{2,7})\\ & = \text{1,8}\text{ dm$^{3}$} \end{align*}

If $\text{4,2}$ $\text{g}$ of magnesium sulfate is dissolved in $\text{350}$ $\text{cm$^{3}$}$ of water, what is the concentration of the solution?

$\text{0,1}$ $\text{mol·dm$^{-3}$}$
$\text{0,05}$ $\text{mol·dm$^{-3}$}$
$\text{0,003}$ $\text{mol·dm$^{-3}$}$
$\text{0,0001}$ $\text{mol·dm$^{-3}$}$

The concentration is: $\text{0,1}$ $\text{mol·dm$^{-3}$}$

The number of moles of magnesium sulfate is: \begin{align*} M_{\text{MgSO}_{4}} &= \text{24,3} + \text{32,1} + \text{16} \times 4 \\ &= \text{120,4 g·mol$^{-1}$} \end{align*} \begin{align*} n&=\frac{m}{M}\\ &=\frac{\text{4,2}}{\text{120,4}} \\ &=\text{0,0348}\ldots\text{ mol} \end{align*}

To convert from $\text{cm$^{3}$}$ to $\text{dm$^{3}$}$ we divide by $\text{1 000}$: \begin{align*} \frac{\text{350}\text{ cm$^{3}$}}{\text{1 000}}& = \text{0,35}\text{ dm$^{3}$} \end{align*}

The concentration is: \begin{align*} c& = \frac{n}{V}\\ & = \frac{\text{0,0348}\ldots}{\text{0,35}}\\ &\approx \text{0,1}\text{ mol·dm$^{-3}$} \end{align*}

Gold is occasionally found as the mineral calaverite. Calaverite contains a telluride of gold ($\text{AuTe}_{2}$). Phumza wants to know how much calaverite is in a sample of rock. She finds that the rock sample weighs $\text{3,6}$ $\text{g}$. After performing some experiments she finds that the mass of calaverite and the crucible is $\text{2,4}$ $\text{g}$. The mass of the crucible is $\text{0,3}$ $\text{g}$. What is the percent purity of calaverite in the sample?

$\text{63}\%$
$\text{54}\%$
$\text{58}\%$
$\text{67}\%$

The percent purity is $\text{58}\%$.

Percent purity is given by:

\[\% \text{purity} = \frac{\text{mass of compound}}{\text{mass of sample}} \times 100\]

We are given the mass of the product and the mass of the crucible. We need to subtract the mass of the crucible from this to obtain just the mass of the compound.

\begin{align*} \text{Mass compound} & = \text{2,4}\text{ g} - \text{0,3}\text{ g} \\ & = \text{2,1}\text{ g} \end{align*}

Substituting the calculated mass into the equation for percent purity gives:

\begin{align*} \% \text{purity} &= \frac{\text{mass of compound}}{\text{mass of sample}} \times 100\\ & = \frac{\text{2,1}}{\text{3,6}} (\text{100})\\ & = \text{58}\% \end{align*}

Write down a balanced equation for the reaction which takes place when these two solutions are mixed.

\[2\text{NaOH} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{Na}_{2}\text{SO}_{4} + 2\text{H}_{2}\text{O}\]

Calculate the number of moles of sulfuric acid which were added to the sodium hydroxide solution.

\begin{align*} c & = \frac{n}{V} \\ \text{0,1} & = \frac{n}{\text{0,3}} \\ n & = \text{0,03}\text{ mol} \end{align*}

$\text{0,03}$ $\text{mol}$ were added.

Is the number of moles of sulfuric acid enough to fully neutralise the sodium hydroxide solution? Support your answer by showing all relevant calculations.

We calculate the number of moles of sodium hydroxide in the solution

\begin{align*} c & = \frac{n}{V} \\ \text{0,5} & = \frac{n}{\text{0,2}} \\ n & = \text{0,1}\text{ mol} \end{align*}

From the molar ratio of sulfuric acid to sodium hydroxide we find that we can neutralise

\begin{align*} n_{\text{NaOH}} & = n_{\text{H}_{2}\text{SO}_{4}} \times \frac{ \text{stoichiometric coefficient NaOH}}{\text{stoichiometric coefficient H}_{2}\text{SO}_{4}}\\ & = \text{0,03}\text{ mol } \text{H}_{2}\text{SO}_{4} \times \frac{\text{2}\text{ mol } \text{NaOH}}{\text{1}\text{ mol } \text{ H}_{2}\text{SO}_{4}}\\ & = \text{0,06}\text{ mol} \end{align*}

of sodium hydroxide. We have $\text{0,1}$ $\text{mol}$ of sodium hydroxide and so the sodium hydroxide will not be fully neutralised.

Given the equation:

\[2\text{NaOH(aq)} + \text{H}_{2}\text{SO}_{4}\text{(aq)} → \text{Na}_{2}\text{SO}_{4}\text{(aq)} + 2\text{H}_{2}\text{O(ℓ)}\]

$\text{25}$ $\text{cm$^{3}$}$ of a $\text{0,7}$ $\text{mol·dm$^{-3}$}$ sulfuric acid ($\text{H}_{2}\text{SO}_{4}$) solution was pipetted into a conical flask and titrated with sodium hydroxide ($\text{NaOH}$). It was found that $\text{23}$ $\text{cm$^{3}$}$ of the sodium hydroxide was needed to neutralise the acid. Calculate the concentration of the sodium hydroxide.

Write down all the information you know about the reaction, and make sure that the equation is balanced.

$\text{H}_{2}\text{SO}_{4}$: $V =\text{25}\text{ cm$^{3}$}$; $c= \text{0,7}\text{ mol·dm$^{-3}$}$

$\text{NaOH}$: $V=\text{23}\text{ cm$^{3}$}$

The equation is already balanced.

Now we can work out the concentration of the sodium hydroxide:

\begin{align*} \frac{c_{a}V_{a}}{n_{a}} & = \frac{c_{b}V_{b}}{n_{b}} \\ \frac{(\text{0,7})(\text{25})}{\text{1}} & = \frac{(c_{\text{NaOH}})(\text{23})}{\text{2}} \\ \text{17,5} & = (\text{11,5})c_{\text{NaOH}}\\ c_{\text{NaOH}} & = \text{1,52}\text{ mol·dm$^{-3}$} \end{align*}

Ozone ($\text{O}_{3}$) reacts with nitrogen monoxide gas ($\text{NO}$) to produce $\text{NO}_{2}$ gas. The $\text{NO}$ gas forms largely as a result of emissions from the exhausts of motor vehicles and from certain jet planes. The $\text{NO}_{2}$ gas also contributes to the brown smog (smoke and fog), which is seen over most urban areas. This gas is also harmful to humans, as it causes breathing (respiratory) problems. The following equation indicates the reaction between ozone and nitrogen monoxide:

\[\text{O}_{3}\text{(g)} + \text{NO (g)} → \text{O}_{2}\text{(g)} + \text{NO}_{2}\text{(g)}\]

In one such reaction $\text{0,74}$ $\text{g}$ of $\text{O}_{3}$ reacts with $\text{0,67}$ $\text{g}$ $\text{NO}$.

Calculate the number of moles of $\text{O}_{3}$ and of $\text{NO}$ present at the start of the reaction.

Moles of $\text{O}_{3}$: \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{0,74}}{\text{48}} \\ & = \text{0,0154}\text{ mol} \end{align*}

Moles of $\text{NO}$: \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{0,67}}{\text{30}} \\ & = \text{0,0223}\text{ mol} \end{align*}

Identify the limiting reagent in the reaction and justify your answer.

Now we look at the number of moles of product that each reactant can form.

The mole ratio of $\text{O}_{3}$ to $\text{NO}_{2}$ is $1:1$. So the number of moles of $\text{NO}_{2}$ that can be produced the available $\text{O}_{3}$ is $\text{0,0154}$ $\text{mol}$

The mole ratio of $\text{NO}$ to $\text{NO}_{2}$ is $1:1$. So the number of moles of $\text{NO}_{2}$ that can be produced from the available $\text{NO}$ is $\text{0,0223}$ $\text{mol}$

Therefore $\text{O}_{3}$ is the limiting reagent.

Calculate the mass of $\text{NO}_{2}$ produced from the reaction.

We have $\text{0,0154}$ $\text{mol}$ of $\text{NO}_{2}$.

The maximum mass of $\text{NO}_{2}$ that can be produced is calculated as follows:

\begin{align*} m & = nM\\ & = (\text{0,0154})(\text{46}) \\ & = \text{0,71}\text{ g} \end{align*}

The maximum amount (theoretical yield) of $\text{NO}_{2}$ that can be produced is $\text{0,71}$ $\text{g}$.

Calcium carbonate decomposes on heating to produce calcium oxide and oxygen according to the following equation:

\[\text{CaCO}_{3}\text{(s)} → \text{CaO (s)} + \text{O}_{2}\text{(g)}\]

Thabang carries out the above reaction using $\text{127}$ $\text{g}$ of calcium carbonate. He finds that he gets $\text{68,2}$ $\text{g}$ of calcium oxide. What is the percentage yield?

There is only one reactant and so we do not need to find the limiting reagent.

We find the number of moles of calcium carbonate: \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{127}}{\text{100}} \\ & = \text{1,27}\text{ mol} \end{align*}

Now we find the number of moles of calcium oxide.

The mole ratio of $\text{CaCO}_{3}$ to $\text{CaO}$ is $1:1$. So the number of moles of $\text{CaO}$ is also $\text{1,27}$ $\text{mol}$.

The maximum mass of calcium oxide that can be produced is:

\begin{align*} m & = nM\\ & = (\text{1,27})(\text{56}) \\ & = \text{71,12}\text{ g} \end{align*}

The maximum amount (theoretical yield) of calcium oxide that can be produced is $\text{71,12}$ $\text{g}$.

The percent yield is:

\begin{align*} \% \text{yield} &= \frac{\text{actual yield}}{\text{theoretical yield}} \times \text{100}\\ & = \frac{\text{68,2}}{\text{71,12}} \times \text{100}\\ & = \text{96,9}\% \end{align*}

Some airbags contain a mixture of sodium azide ($\text{NaN}_{3}$) and potassium nitrate ($\text{KNO}_{3}$). When a car crash is detected by the signalling system, the sodium azide is heated until it decomposes to form nitrogen gas and sodium metal:

\[2\text{NaN}_{3}\text{(s)} \rightarrow 2\text{Na (s)} + 3\text{N}_{2}\text{(g)}\]

The potassium nitrate then reacts with the sodium metal forming more nitrogen:

\[10\text{Na (s)} + 2\text{KNO}_{3}\text{(s)} \rightarrow \text{K}_{2} \text{O (s)} + 5\text{Na}_{2}\text{O (s)} + \text{N}_{2}\text{(g)}\]

A typical passenger side airbag contains $\text{250}$ $\text{g}$ of sodium azide.

What mass of sodium metal is formed in the first reaction?

To find the amount of sodium metal we first calculate how many moles of sodium azide is in the airbag:

\begin{align*} n & = \frac{\text{250}}{\text{65}} \\ & = \text{3,846}\text{ mol} \end{align*}

Next we find the number of moles of sodium. The molar ratio of sodium azide to sodium is $2:2$ (or $1:1$) so the number of moles of sodium is: $\text{3,846}$ $\text{mol}$

And the mass of sodium metal is:

\begin{align*} m & = (\text{3,846})(\text{23}) \\ & = \text{88,461}\text{ g} \end{align*}

What is the total volume of nitrogen gas formed from both reactions?

To find the total volume of nitrogen produced we need to calculate the volume of nitrogen produced in each reaction and then add these two numbers together.

For the first reaction we have $\text{3,846}$ $\text{mol}$ of sodium azide. The mole ratio of sodium azide to nitrogen is $2:3$, so the number of moles of nitrogen is:

\begin{align*} n_{\text{N}_{2}} & = n_{\text{NaN}_{3}} \times \frac{\text{stoichiometric coefficient N}_{2}}{\text{stoichiometric coefficient NaN}_{3}}\\ & = \text{3,846}\text{ mol} \text{NaN}_{3} \times \frac{\text{2}\text{ mol} \text{N}_{2}}{\text{3}\text{ mol} \text{ NaN}_{3}}\\ & = \text{2,564}\text{ mol} \text{N}_{2} \end{align*}

Now we find the volume:

\begin{align*} V & = \text{22,4}n \\ & = (\text{22,4})(\text{2,564}) \\ & = \text{57,436}\text{ dm$^{3}$} \end{align*}

For the second reaction we have $\text{3,846}$ $\text{mol}$ of sodium (the sodium from the first reaction is used up in the second reaction). The mole ratio of sodium to nitrogen is $10:1$, so the number of moles of nitrogen is:

\begin{align*} n_{\text{N}_{2}} & = n_{\text{Na}} \times \frac{\text{stoichiometric coefficient N}_{2}}{\text{stoichiometric coefficient Na}}\\ & = \text{3,846}\text{ mol} \text{Na} \times \frac{\text{1}\text{ mol} \text{N}_{2}}{\text{10}\text{ mol} \text{ Na}}\\ & = \text{0,385}\text{ mol} \text{N}_{2} \end{align*}

Now we find the volume:

\begin{align*} V & = \text{22,4}n \\ & = (\text{22,4})(\text{0,385}) \\ & = \text{8,615}\text{ dm$^{3}$} \end{align*}

And the total volume of nitrogen is:

\begin{align*} V_{T} & = V_{1} + V_{2} \\ & = \text{57,436} + \text{8,615} \\ & = \text{66,051}\text{ dm$^{3}$} \end{align*}

How much potassium nitrate (in $\text{g}$) is needed for all the sodium to be used up in the second reaction?

The number of moles of sodium is: $\text{3,846}$ $\text{mol}$. The molar ratio of sodium to potassium nitrate is: $10:2$ or $5:1$. So the number of moles of potassium nitrate is:

\begin{align*} n_{\text{KNO}_{3}} & = n_{\text{Na}} \times \frac{\text{stoichiometric coefficient KNO}_{3}}{\text{stoichiometric coefficient Na}}\\ & = \text{3,846}\text{ mol} \text{Na} \times \frac{\text{1}\text{ mol} \text{KNO}_{3}}{\text{5}\text{ mol} \text{ Na}}\\ & = \text{0,7692}\text{ mol} \end{align*}

And the mass of potassium nitrate needed is:

\begin{align*} m & = (\text{0,7692})(\text{101}) \\ & = \text{77,69}\text{ g} \end{align*}

To ensure a complete reaction we could use $\text{78}$ $\text{g}$ of potassium nitrate.

Chlorofluorocarbons (CFC's) are a class of compounds that have a long history of use in refrigerators. CFC's are slowly being phased out as they deplete the amount of ozone in the ozone layer. Jabu has a sample of a CFC that has the following percentage composition: $\text{14,05}\%$ carbon, $\text{41,48}\%$ chlorine and $\text{44,46}\%$ fluorine.

Determine the molecular formula of this CFC if the molar mass is $\text{171}$ $\text{g·mol$^{-1}$}$.

In $\text{100}$ $\text{g}$ of the CFC, there is: $\text{14,05}$ $\text{g}$ $\text{C}$, $\text{41,48}$ $\text{g}$ $\text{Cl}$, and $\text{44,46}$ $\text{g}$ $\text{F}$.

$n = \frac{m}{M}$

\begin{align*} n_{\text{C}} & = \frac{\text{14,05}}{\text{12}} = \text{1,17}\text{ mol}\\ n_{\text{Cl}} & = \frac{\text{41,48}}{\text{35}} = \text{1,19}\text{ mol}\\ n_{\text{F}} & = \frac{\text{44,46}}{\text{19}} = \text{2,34}\text{ mol} \end{align*}

To find the empirical formula we first note how many moles of each element we have. Then we divide by the smallest number to get the ratios of each element. This ratio is rounded off to the nearest whole number.

$\text{C}$	$\text{Cl}$	$\text{F}$
$\text{1,17}$	$\text{1,19}$	$\text{2,34}$
$\dfrac{\text{1,17}}{\text{1,17}}$	$\dfrac{\text{1,19}}{\text{1,17}}$	$\dfrac{\text{2,34}}{\text{1,17}}$
1	1	2

The empirical formula is $\text{CClF}_{2}$.

The molar mass of the CFC using the empirical formula is $\text{85}$ $\text{g·mol$^{-1}$}$. However the question gives the molar mass as $\text{171}$ $\text{g·mol$^{-1}$}$. We divide the given molar mass by the calculated molar mass to find the molecular formula: $\frac{\text{171}}{\text{85}} = \text{2}$. Therefore the molecular formula is: $\text{C}_{2}\text{Cl}_{2}\text{F}_{4}$.

A sample containing tin dioxide ($\text{SnO}_{2}$) is to be tested to see how much tin dioxide it contains. The sample weighs $\text{6,2}$ $\text{g}$. Sulfuric acid ($\text{H}_{2}\text{SO}_{4}$) is added to the sample and tin sulfate ($\text{Sn}(\text{SO}_{4})_{2}$) forms. The equation for this reaction is: \begin{align*} \text{SnO}_{2}\text{(s)} + 2\text{H}_{2}\text{SO}_{4}\text{(aq)} \rightarrow \text{Sn(SO}_{4}\text{)}_{2}\text{(s)} + 2\text{H}_{2}\text{O(ℓ)} \end{align*}

If the mass of tin sulfate produced is $\text{4,7}$ $\text{g}$, what is the percent purity of the sample?

The number of moles of tin sulfate is (the molar mass of tin is ($\text{119}$ $\text{g·mol$^{-1}$}$)): \begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{6,2}}{\text{311}} \\ & = \text{0,0199}\text{ mol} \end{align*}

The molar ratio of tin sulfate to tin dioxide is 1:1. Therefore the number of moles of tin dioxide is $\text{0,0199}$ $\text{mol}$.

The mass of tin dioxide is: \begin{align*} m & = nM \\ & = (\text{0,0199})(\text{151}) \\ & = \text{3,01}\text{ g} \end{align*}

Substituting the calculated mass into the equation for percent purity gives: \begin{align*} \% \text{purity} &= \frac{\text{mass of compound}}{\text{mass of sample}} \times 100\\ & = \frac{\text{3,01}}{\text{6,2}} \times (100)\\ & = \text{48,6}\% \end{align*}

If the volume of methane used is $\text{4}$ $\text{dm$^{3}$}$, what volume of carbon monoxide and hydrogen will be produced?

The volume of carbon monoxide produced is:

\begin{align*} V_{\text{CO}_{2}} & = \frac{a}{b}V_{\text{CH}_{4}}\\ & = \frac{1}{1}(\text{4})\\ & = \text{4}\text{ dm$^{3}$} \end{align*}

The volume of hydrogen produced is:

\begin{align*} V_{\text{H}_{2}} & = \frac{a}{b}V_{\text{CH}_{4}}\\ & = \frac{1}{3}(\text{4})\\ & = \text{1,33}\text{ dm$^{3}$} \end{align*}

Will this amount of methane produce the correct mixture?

Yes. Since the two gases have a ratio of $1:3$ we will always have three times as much carbon monoxide as hydrogen gas.

8.4 Chapter summary

Table of Contents

9.1 Introduction

\(\text{C}\)	\(\text{Cl}\)	\(\text{F}\)
\(\text{1,17}\)	\(\text{1,19}\)	\(\text{2,34}\)
\(\dfrac{\text{1,17}}{\text{1,17}}\)	\(\dfrac{\text{1,19}}{\text{1,17}}\)	\(\dfrac{\text{2,34}}{\text{1,17}}\)
1	1	2