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*}

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*}