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Gases And Solutions

Chapter 8: Quantitative aspects of chemical change

This topic builds on the gr10 chapter of the same title. A quick revision (no more than 15 minutes) of the gr10 work would greatly help learners to recall how to calculate molar masses, number of moles and concentration. In this chapter learners will learn about molar volumes of gases, concentration of solutions, more on stoichiometry and volume relationships in gaseous reactions. Topics covered in this chapter are:

  • 1 mole of gas occupies \(\text{22,4}\) \(\text{dm$^{3}$}\) and expansion to any number of moles of gas

    In grade 10 learners learnt about the molar volume of gases. This concept is refreshed here and then expanded on to cover any number of moles of gases. This is a very important result that is used in stoichiometric calculations to go from volume of a gas to number of moles and from there via the balanced equation to the desired quantity of one of the products or reactants in that reaction.

  • Interpretation of chemical equations

    Learners should be able to understand that chemical equations can be interpreted in words or in symbols (as seen in grade 10) or in terms of the quantities of the substances involved in the reaction. This second interpretation is a valuable skill for stoichiometric calculations.

  • Calculation of molar concentration of solutions and expanding this to titration calculations

    CAPs lists an activity on titration calculations which is worked into the text with concentration calculations. In grade 10 learners learnt about calculating concentration, in this chapter concentration calculations are expanded to cover titrations.

  • Limiting reagents

    In grade 10 learners saw how to use the balanced chemical equation to move from products to reactants and vice versa. This is now expanded on to include information about both reactants where learners have to determine which reactant is limiting and which is in excess before calculating the amount of product formed.

  • Percent yield

    In this section learners are introduced to percent yield. Care must be taken in this section to ensure that learners do not mix up theoretical yield and percent yield. The percent yield can only be determined from experiment, while the theoretical yield comes from stoichiometric calculations.

  • Empirical and molecular formula

    This is revision from grade 10. Learners are reminded how to calculate the empirical and molecular formula of a substance and are reminded about percent composition.

  • Percent purity

    The concept of percent purity is the last concept that is covered. This looks at how pure a sample of a compound is. It is important to note that CAPs refers to percent purity as being interchangeable with percent composition which is not strictly true. Percent composition deals with how much of a particular element is in a substance while percent purity deals with how much of a compound is in a sample or how pure your final product in a reaction is.

  • Calculations involving gases, liquids and solids

    CAPs also lists an activity on precipitation calculations which can be included in the final topic.

    At the end of the chapter learners bring together all the concepts that they have learnt into this topic. Learners are introduced to the idea that the balanced chemical equation can be used to calculate any quantity of a substance from any of the other products or reactions in that reaction. This is particularly important as reactions do not occur in just one state but occur in multiple states and being able to move freely between, for example, calculating moles of a gas to calculating mass of a substance, is an important skill.

Wherever we look in real life we see the importance of mixing things in precise quantities. Cooking and baking, the medicines you take when sick, the products that you buy, all of these rely on the ingredients being mixed in specific amounts. And even the amount of product formed relies on how much of each ingredient is used. In this chapter we will look at some of these quantities and how they can be calculated.

In grade 10 we learnt about writing chemical equations and about the information that can be obtained from a balanced chemical equation. In this chapter we are going to explore these concepts further and learn more about gases, solutions and reactions. We will explore the concept of theoretical yield in greater detail and learn about limiting reagents.

  • Ratio and proportion - Physical Sciences, Grade 10, Science skills
  • Equations - Mathematics, Grade 10, Equations and inequalities
  • Units and unit conversions - Physical Sciences, Grade 10, Science skills

8.1 Gases and solutions (ESBP4)

We will begin by taking a closer look at gases and solutions and work out how to solve problems relating to them.

Molar volumes of gases (ESBP5)

It is possible to calculate the volume of one mole of gas at standard temperature and pressure (STP) using what we now know about gases.

STP is a temperature of \(\text{273}\) \(\text{K}\) and a pressure of \(\text{101,3}\) \(\text{kPa}\). The amount of gas is usually \(\text{1}\) \(\text{mol}\).

We write down all the values that we know about one mole of gas at STP:

\begin{align*} p & = \text{101,3}\text{ kPa} = \text{101 300}\text{ Pa} \\ n & = \text{1}\text{ mol} \\ R & = \text{8,31}\text{ J·K$^{-1}$·mol$^{-1}$} \\ T & = \text{273}\text{ K} \end{align*}

Now we can substitute these values into the ideal gas equation:

\begin{align*} pV & = nRT \\ (\text{101 300})V & = (\text{1})(\text{8,31})(\text{273}) \\ (\text{101 300})V & = \text{2 265,9} \\ V & = \text{0,0224}\text{ m$^{3}$} \\ V & = \text{22,4}\text{ dm$^{3}$} \end{align*}

The volume of 1 mole of gas at STP is \(\text{22,4}\) \(\text{dm$^{3}$}\).

And if we had any number of moles of gas, not just one mole then we would get: \[\boxed{V_{g} = \text{22,4}n_{g}}\]

The standard units used for this equation are p in \(\text{Pa}\), V in \(\text{m$^{3}$}\) and T in \(\text{K}\). Remember also that \(\text{1 000}\) \(\text{cm$^{3}$}\) \(=\) \(\text{1}\) \(\text{dm$^{3}$}\) and \(\text{1 000}\) \(\text{dm$^{3}$}\) \(=\) \(\text{1}\) \(\text{m$^{3}$}\).

Worked example 1: Molar gas volume

What is the volume of \(\text{2,3}\) \(\text{mol}\) of hydrogen gas at STP?

Find the volume

\begin{align*} V_{g} & = (\text{22,4})n_{g} \\ & = (\text{22,4})(\text{2,3})\\ & = \text{51,52}\text{ dm$^{3}$} \end{align*}

Reactions and gases (ESBP6)

Some reactions take place between gases. For these reactions we can work out the volumes of the gases using the fact that volume is proportional to the number of moles.

We can use the following formula:

\begin{align*} V_{\text{A}} & = \frac{a}{b} V_{\text{B}} \end{align*}

where:

\begin{align*} V_{A} & = \text{volume of A}\\ V_{B} & = \text{volume of B}\\ a & = \text{stoichiometric coefficient of A}\\ b & = \text{stoichiometric coefficient of B} \end{align*}

The number in front of a reactant or a product in a balanced chemical equation is called the stoichiometric coefficient or stoichiometric ratio.

Worked example 2: Volume and gases

Hydrogen and oxygen react to form water according to the following equation:

\[2\text{H}_{2} \text{(g)} + \text{O}_{2} \text{(g)} \rightarrow 2\text{H}_{2}\text{O} \text{(g)}\]

If \(\text{3}\) \(\text{dm$^{3}$}\) of oxygen is used, what volume of water is produced?

Determine the volume of water produced in the reaction.

We use the equation given above to work out the volume of water needed: \begin{align*} V_{A} & = \frac{a}{b}V_{B} \\ V_{\text{H}_{2}\text{O}} & = \frac{2}{1}V_{\text{O}_{2}} \\ & = 2(\text{3})\\ & = \text{6}\text{ dm$^{3}$} \end{align*}

We can interpret the chemical equation in the worked example above (\(2\text{H}_{2} \text{(g)} + \text{O}_{2} \text{(g)} \rightarrow 2\text{H}_{2}\text{O} \text{(g)}\)) as:

\(\text{2}\) moles of hydrogen react with \(\text{1}\) mole of oxygen to produce \(\text{2}\) moles of water. We can also say that \(\text{2}\) volumes of hydrogen react with \(\text{1}\) volume of oxygen to produce \(\text{2}\) volumes of water.

Worked example 3: Gas phase calculations

What volume of oxygen at STP is needed for the complete combustion of \(\text{3,3}\) \(\text{dm$^{3}$}\) of propane (\(\text{C}_{3}\text{H}_{8}\))? (Hint: \(\text{CO}_{2}\) and \(\text{H}_{2}\text{O}\) are the products as in all combustion reactions)

Write a balanced equation for the reaction.

\(\text{C}_{3}\text{H}_{8}\text{(g)} + 5\text{O}_{2}\text{(g)} → 3\text{CO}_{2}\text{(g)} + 4\text{H}_{2}\text{O (g)}\)

Determine the volume of oxygen needed for the reaction.

We use the equation given above to work out the volume of oxygen needed: \begin{align*} V_{A} & = \frac{a}{b}V_{B} \\ V_{\text{O}_{2}} & = \frac{5}{1}V_{\text{C}_{3}\text{H}_{8}} \\ & = 5(\text{3,3})\\ & = \text{16,5}\text{ dm$^{3}$} \end{align*}

Exercise 8.1

Methane burns in oxygen, forming water and carbon dioxide according to the following equation:

\[\text{CH}_{4} \text{(g)} + 2\text{O}_{2} \text{(g)} \rightarrow 2\text{H}_{2}\text{O} \text{(g)} + \text{CO}_{2}\text{(g)}\]

If \(\text{4}\) \(\text{dm$^{3}$}\) of methane is used, what volume of water is produced?

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

Solutions (ESBP7)

In grade 10 you learnt how to calculate the molar concentration of a solution. The molar concentration of a solution is the number of moles of solute per litre of solvent (\(\text{mol·L$^{-1}$}\)). This is more commonly given as moles of solute per cubic decimetre of solution (\(\text{mol·dm$^{-3}$}\)).

To calculate concentration we use \(C = \frac{n}{V}\), where \(C\) is the molar concentration, \(n\) is the number of moles and \(V\) is the volume of the solution.

Calculating molar concentrations is useful to determine how much solute we need to add to a given volume of solvent in order to make a standard solution.

A standard solution is a solution in which the concentration is known to a high degree of precision. When we work with standard solutions we can take the concentration to be constant.

When you are busy with these calculations, you will need to remember the following:

\(\text{1}\) \(\text{dm$^{3}$}\)=\(\text{1}\) \(\text{L}\)=\(\text{1 000}\) \(\text{mL}\)=\(\text{1 000}\) \(\text{cm$^{3}$}\), therefore dividing a volume in \(\text{cm$^{3}$}\) by \(\text{1 000}\) will give you the equivalent volume in \(\text{dm$^{3}$}\).

Worked example 4: Concentration calculations

How much sodium chloride (in g) will one need to prepare \(\text{500}\) \(\text{cm$^{3}$}\) of a standard solution with a concentration of \(\text{0,01}\) \(\text{mol·dm$^{-3}$}\)?

Convert all quantities into the correct units for this equation.

The volume must be converted to \(\text{dm$^{3}$}\): \begin{align*} V & = \frac{\text{500}}{\text{1 000}} \\ & = \text{0,5}\text{ dm$^{3}$} \end{align*}

Calculate the number of moles of sodium chloride needed.

\begin{align*} C &= \frac{n}{V} \\ \text{0,01} & = \frac{n}{\text{0,5}} \\ n & = \text{0,005}\text{ mol} \end{align*}

Convert moles of \(\text{NaCl}\) to mass.

To find the mass of \(\text{NaCl}\) we need the molar mass of \(\text{NaCl}\). We can get this from the periodic table (recall from grade 10 how to calculate the molar mass of a compound).

\begin{align*} m & = nM \\ & = (\text{0,005})(\text{58})\\ & = \text{0,29}\text{ g} \end{align*}

The mass of sodium chloride needed is \(\text{0,29}\) \(\text{g}\)

We will now look at another use of concentration which is for titration calculations.

Titrations

A titration is a technique for determining the concentration of an unknown solution. Titrations can be done using many different types of reactions. Acid-base reactions and redox reactions are both commonly used for titrations.

In grade 10 you did a simple acid-base titration. Now we will look at how to calculate the concentration of an unknown solution using an acid-base titration.

When performing a titration we say that the substance of unknown concentration is titrated with the standard solution. A pipette is a measuring device that is used to measure an exact amount of a liquid. If you use a pipette to add liquid to a flask then you would say that the liquid was pipetted into a flask.

We can reduce the number of calculations that we have to do in titration calculations by using the following: \begin{align*} \frac{C_{A}V_{A}}{a} & = \frac{C_{B}V_{B}}{b} \end{align*}

The \(a\) and \(b\) are the stoichiometric coefficients of compounds A and B respectively.

Worked example 5: Titration calculation

Given the equation:

\(\text{NaOH (aq)} + \text{HCl (aq)} → \text{NaCl (aq)} + \text{H}_{2}\text{O (l)}\)

\(\text{25}\) \(\text{cm$^{3}$}\) of a \(\text{0,2}\) \(\text{mol·dm$^{-3}$}\) hydrochloric acid solution was pipetted into a conical flask and titrated with sodium hydroxide. It was found that \(\text{15}\) \(\text{cm$^{3}$}\) 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.

\(\text{NaOH}\): \(V =\text{15}\text{ cm$^{3}$}\)

\(\text{HCl}\): \(V=\text{25}\text{ cm$^{3}$}\); \(C= \text{0,2}\text{ mol·dm$^{-3}$}\)

The equation is already balanced.

Convert the volume to \(\text{dm$^{3}$}\)

\begin{align*} V_{\text{NaOH}} & = \frac{\text{15}}{\text{1 000}} \\ & = \text{0,015}\text{ dm$^{3}$} \end{align*}\begin{align*} V_{\text{HCl}} & = \frac{\text{25}}{\text{1 000}} \\ & = \text{0,025}\text{ dm$^{3}$} \end{align*}

Calculate the concentration of the sodium hydroxide

\begin{align*} \frac{C_{A}V_{A}}{a} & = \frac{C_{B}V_{B}}{b} \\ \frac{(\text{0,2})(\text{0,025})}{\text{1}} & = \frac{(C_{\text{NaOH}})(\text{0,015})}{\text{1}} \\ \text{0,005} & = (\text{0,015})C_{\text{NaOH}}\\ C_{\text{NaOH}} & = \text{0,33}\text{ mol·dm$^{-3}$} \end{align*}

The concentration of the \(\text{NaOH}\) solution is \(\text{0,33}\) \(\text{mol·dm$^{-3}$}\)

Worked example 6: Titration calculation

\(\text{4,9}\) \(\text{g}\) of sulfuric acid is dissolved in water and the final solution has a volume of \(\text{220}\) \(\text{cm$^{3}$}\). Using an acid-base titration, it was found that \(\text{20}\) \(\text{cm$^{3}$}\) of this solution was able to completely neutralise \(\text{10}\) \(\text{cm$^{3}$}\) of a sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in \(\text{mol·dm$^{-3}$}\).

Write a balanced equation for the titration reaction.

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

Calculate the concentration of the sulfuric acid solution.

First convert the volume into \(\text{dm$^{3}$}\):

\begin{align*} V & = \frac{\text{220}}{\text{1 000}} = \text{0,22}\text{ dm$^{3}$} \end{align*}

Then calculate the number of moles of sulfuric acid:

\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{4,9}}{\text{98}}\\ & = \text{0,05}\text{ mol} \end{align*}

Now we can calculate the concentration of the sulfuric acid:

\begin{align*} C & = \frac{n}{V} \\ & = \frac{\text{0,05}}{\text{0,22}}\\ & = \text{0,227}\text{ mol·dm$^{-3}$} \end{align*}

Calculate the concentration of the sodium hydroxide solution.

Remember that only \(\text{20}\) \(\text{cm$^{3}$}\) or \(\text{0,02}\) \(\text{dm$^{3}$}\) of the sulfuric acid solution is used.

\begin{align*} \frac{C_{1}V_{1}}{n_{1}} & = \frac{C_{2}V_{2}}{n_{2}} \\ \frac{(\text{0,227})(\text{0,02})}{\text{1}} & = \frac{(C_{\text{NaOH}})(\text{0,01})}{\text{2}} \\ \text{0,00454} & = (\text{0,005})C_{\text{NaOH}}\\ C_{\text{NaOH}} & = \text{0,909}\text{ mol·dm$^{-3}$} \end{align*}

Gases and solutions

Exercise 8.2

Acetylene (\(\text{C}_{2}\text{H}_{2}\)) burns in oxygen according to the following reaction:

\begin{align*} 2\text{C}_{2}\text{H}_{2}\text{(g)} + 5\text{O}_{2}\text{(g)} \rightarrow 4\text{CO}_{2}\text{(g)} + 2\text{H}_{2}\text{O (g)} \end{align*}

If \(\text{3,5}\) \(\text{dm$^{3}$}\) of acetylene gas is burnt, what volume of carbon dioxide will be produced?

\begin{align*} V_{A} & = \frac{a}{b}V_{B} \\ V_{\text{CO}_{2}} & = \frac{4}{2}(\text{3,5}) \\ & = \text{7}\text{ dm$^{3}$} \end{align*}

\(\text{130}\) \(\text{g}\) of magnesium chloride (\(\text{MgCl}_{2}\)) is dissolved in \(\text{300}\) \(\text{mL}\) of water.

Calculate the concentration of the solution.

The number of moles of magnesium chloride is: \begin{align*} n&=\frac{m}{M}\\ &=\frac{\text{130}}{\text{95}} \\ &=\text{1,37}\text{ mol} \end{align*}

To convert from \(\text{mL}\) to \(\text{dm$^{3}$}\) we divide by \(\text{1 000}\) (\(\text{1 000}\) \(\text{mL}\) = \(\text{1}\) \(\text{L}\) = \(\text{1}\) \(\text{dm$^{3}$}\)): \begin{align*} \frac{\text{300}\text{ mL}}{\text{1 000}}& = \text{0,3}\text{ dm$^{3}$} \end{align*}

The concentration is:

\begin{align*} C& = \frac{n}{V}\\ & = \frac{\text{1,37}}{\text{0,3}}\\ &= \text{4,56}\text{ mol·dm$^{-3}$} \end{align*}

What mass of magnesium chloride would need to be added for the concentration to become \(\text{6,7}\) \(\text{mol·dm$^{-3}$}\)?

We first work out what mass of magnesium chloride is needed to make a solution with a concentration of \(\text{6,7}\) \(\text{mol·dm$^{-3}$}\). The volume will still be \(\text{0,3}\) \(\text{dm$^{3}$}\). \begin{align*} n & = CV\\ & = (\text{6,7})(\text{0,3})\\ &= \text{2,01}\text{ mol} \end{align*}

This gives a mass of: \begin{align*} m&= nM\\ &= (\text{2,01})(\text{95}) \\ &=\text{190,95}\text{ g} \end{align*}

To get from a concentration of \(\text{4,56}\) \(\text{mol·dm$^{-3}$}\) to a concentration of \(\text{6,7}\) \(\text{mol·dm$^{-3}$}\) we must add: \begin{align*} \text{190,95}-\text{130} & = \text{60,95}\text{ g} \end{align*} of magnesium chloride.

Given the equation:

\(\text{KOH (aq)} + \text{HNO}_{3}\text{(aq)} → \text{KNO}_{3}\text{(aq)} + \text{H}_{2}\text{O (l)}\)

\(\text{20}\) \(\text{cm$^{3}$}\) of a \(\text{1,3}\) \(\text{mol·dm$^{-3}$}\) potassium hydroxide (\(\text{KOH}\)) solution was pipetted into a conical flask and titrated with nitric acid (\(\text{HNO}_{3}\)). It was found that \(\text{17}\) \(\text{cm$^{3}$}\) of the nitric acid was needed to neutralise the base. Calculate the concentration of the nitric acid.

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

\(\text{KOH}\): \(V =\text{20}\text{ cm$^{3}$}\); \(C= \text{1,3}\text{ mol·dm$^{-3}$}\)

\(\text{HNO}_{3}\): \(V=\text{17}\text{ cm$^{3}$}\)

The equation is already balanced.

Next convert the volumes to \(\text{dm$^{3}$}\)

\begin{align*} V_{\text{KOH}} & = \frac{\text{20}}{\text{1 000}} \\ & = \text{0,020}\text{ dm$^{3}$} \end{align*} \begin{align*} V_{\text{HNO}_{3}} & = \frac{\text{17}}{\text{1 000}} \\ & = \text{0,017}\text{ dm$^{3}$} \end{align*}

And now we can work out the concentration of the nitric acid:

\begin{align*} \frac{C_{1}V_{1}}{n_{1}} & = \frac{C_{2}V_{2}}{n_{2}} \\ \frac{(\text{1,3})(\text{0,017})}{\text{1}} & = \frac{(C_{\text{HNO}_{3}})(\text{0,020})}{\text{2}} \\ \text{0,0221} & = (\text{0,020})C_{\text{HNO}_{3}}\\ C_{\text{HNO}_{3}} & = \text{1,105}\text{ mol·dm$^{-3}$} \end{align*}

Given the equation:

\(3\text{Ca(OH)}_{2}\text{(aq)} + 2\text{H}_{3}\text{PO}_{4}\text{(aq)} → \text{Ca}_{3}\text{(PO}_{4}\text{)}_{2}\text{(aq)} + 6\text{H}_{2}\text{O (l)}\)

\(\text{10}\) \(\text{cm$^{3}$}\) of a \(\text{0,4}\) \(\text{mol·dm$^{-3}$}\) calcium hydroxide (\(\text{Ca}(\text{OH})_{2}\)) solution was pipetted into a conical flask and titrated with phosphoric acid (\(\text{H}_{3}\text{PO}_{4}\)). It was found that \(\text{11}\) \(\text{cm$^{3}$}\) of the phosphoric acid was needed to neutralise the base. Calculate the concentration of the phosphoric acid.

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

\(\text{Ca}(\text{OH})_{2}\): \(V =\text{10}\text{ cm$^{3}$}\); \(C= \text{0,4}\text{ mol·dm$^{-3}$}\)

\(\text{H}_{3}\text{PO}_{4}\): \(V=\text{11}\text{ cm$^{3}$}\)

The equation is already balanced.

Next convert the volumes to \(\text{dm$^{3}$}\)

\begin{align*} V_{\text{Ca(OH)}_{2}} & = \frac{\text{10}}{\text{1 000}} \\ & = \text{0,010}\text{ dm$^{3}$} \end{align*} \begin{align*} V_{\text{H}_{3}\text{PO}_{4}} & = \frac{\text{11}}{\text{1 000}} \\ & = \text{0,011}\text{ dm$^{3}$} \end{align*}

And now we can work out the concentration of the nitric acid:

\begin{align*} \frac{C_{1}V_{1}}{n_{1}} & = \frac{C_{2}V_{2}}{n_{2}} \\ \frac{(\text{0,4})(\text{0,010})}{\text{3}} & = \frac{(C_{\text{H}_{3}\text{PO}_{4}})(\text{0,011})}{\text{2}} \\ \text{0,001333} & = (\text{0,0055})C_{\text{H}_{3}\text{PO}_{4}}\\ C_{\text{H}_{3}\text{PO}_{4}} & = \text{0,24}\text{ mol·dm$^{-3}$} \end{align*}

A \(\text{3,7}\) \(\text{g}\) sample of an antacid (which contains mostly calcium carbonate) is dissolved in water. The final solution has a volume of \(\text{500}\) \(\text{mL}\). \(\text{25}\) \(\text{mL}\) of this solution is then pipetted into a conical flask and titrated with hydrochloric acid. It is found that \(\text{20}\) \(\text{mL}\) of the hydrochloric acid completely neutralises the antacid solution. What is the concentration of the hydrochloric acid?

The equation for this reaction is:

\[\text{CaCO}_{3}\text{(aq)} + 2\text{HCl (aq)} \rightarrow \text{CaCl}_{2}\text{(aq)} + \text{H}_{2}\text{O (l)} + \text{CO}_{2}\text{(g)}\]

First convert the volume into \(\text{dm$^{3}$}\): \begin{align*} V & = \frac{\text{500}}{\text{1 000}} = \text{0,5}\text{ L}\\ & = \text{0,5}\text{ dm$^{3}$} \end{align*}

Then calculate the number of moles of calcium carbonate: \begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{3,7}}{\text{100}}\\ & = \text{0,037}\text{ mol} \end{align*}

Now we can calculate the concentration of the calcium carbonate: \begin{align*} C & = \frac{n}{V} \\ & = \frac{\text{0,037}}{\text{0,5}}\\ & = \text{0,074}\text{ mol·dm$^{-3}$} \end{align*}

Now calculate the concentration of the hydrochloric acid.

Remember that only \(\text{25}\) \(\text{mL}\) or \(\text{0,025}\) \(\text{dm$^{3}$}\) of the calcium carbonate solution is used.

\begin{align*} \frac{C_{1}V_{1}}{n_{1}} & = \frac{C_{2}V_{2}}{n_{2}} \\ \frac{(\text{0,074})(\text{0,025})}{\text{1}} & = \frac{(C_{\text{HCl}})(\text{0,02})}{\text{2}} \\ \text{0,00185} & = (\text{0,01})C_{\text{HCl}}\\ C_{\text{HCl}} & = \text{0,185}\text{ mol·dm$^{-3}$} \end{align*}