Question

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(3.5 \times 4.5 \times 2.5 \mathrm{~m}\). The density of air at \(26^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3} .(\mathbf{b})\) If the HCN is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon \(^{\text {- }}\) or Acrilan \(^{\circledast}\) burn. Acrilan \(^{\circledast}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is \(50.9 \%\) of the formula by mass. A rug measures \(3.5 \times 4.5 \mathrm{~m}\) and contains \(850 \mathrm{~g}\) of Acrilan \(^{\circledast}\) fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.

Short answer

In a small laboratory room measuring \(3.5 \times 4.5 \times 2.5 \mathrm{~m}\), the lethal dose of HCN is approximately \(2.10 \mathrm{~g}\). To produce this lethal dose, a mass of about \(3.82 \mathrm{~g}\) of NaCN is required. When the rug containing Acrilan fibers burns, it generates approximately \(0.43 \mathrm{~g}\) of HCN, which is less than the lethal dose, so a lethal dose of HCN isn't generated.

Step-by-step solution

1. Calculate the volume of the room

First, let's find the volume of the laboratory room by multiplying its dimensions: \(V = l \cdot w \cdot h = 3.5 \mathrm{~m} \cdot 4.5 \mathrm{~m} \cdot 2.5 \mathrm{~m}\).

2. Convert the volume to cubic centimeters

Next, we need to convert the volume of the room to cubic centimeters so it matches the density of air provided: \(V = V \cdot (100 \mathrm{~cm / m})^3\).

3. Calculate the mass of air in the room

Now we can find the mass of air in the room using the density of air formula: \(m_\text{air} = V \cdot \rho_\text{air}\), where \(\rho_\text{air} = 0.00118 \mathrm{~g} / \mathrm{cm}^3\).

4. Calculate the lethal dose of HCN in the room

Using the lethal dose of HCN per kilogram of air, we can find the lethal dose of HCN in the room: \(\mathrm{Lethal\:dose\:of\:HCN} = m_\text{air} \cdot \frac{300 \mathrm{mg\:HCN}}{1\mathrm{kg\:air}} \). Convert the lethal dose to grams: \(300\mathrm{mg}\) = \(0.3 \mathrm{g}\). #Part b#

1. Determine the stoichiometry of the reaction

From the balanced chemical equation, we can see that 2 moles of NaCN react to form 2 moles of HCN: \(\mathrm{2\:NaCN\:to\:2\:HCN}\).

2. Calculate the moles of HCN for the lethal dose

Using the lethal dose of HCN found in part (a), we can calculate the moles of HCN needed for the lethal dose: \(n_\text{HCN} = \frac{m_\text{HCN}}{M_\text{HCN}}\), where \(M_\text{HCN} = 27.03\,\mathrm{g/mol}\).

3. Calculate the moles and mass of NaCN for the lethal dose

Using the stoichiometry of the reaction, we can calculate the moles of NaCN needed for the lethal dose: \(n_\text{NaCN} = n_\text{HCN}\). Then, we can find the mass of NaCN required using the molar mass of NaCN, \(M_\text{NaCN} = 49.01\,\mathrm{g/mol} \): \(m_\text{NaCN} = n_\text{NaCN} \cdot M_\text{NaCN}\). #Part c#

1. Calculate the mass of HCN in the fibers of the rug

Using the given mass of Acrilan fibers, we can calculate the mass of HCN present in the fibers: \(m_\text{HCN} = 850 \mathrm{~g} \cdot 0.509\), where \(0.509\) is the fraction of HCN by mass in Acrilan.

2. Determine the mass of HCN generated when the rug burns

Now we can find the mass of HCN generated when the rug burns, taking into account the yield and the amount of carpet consumed: \(m_\text{HCN,generated} = m_\text{HCN} \cdot 0.20 \cdot 0.50\), where \(0.20\) is the yield of HCN and \(0.50\) is the fraction of carpet consumed.

3. Compare the mass of HCN generated to the lethal dose

Finally, compare the mass of HCN generated to the lethal dose found in part (a). If \(m_\text{HCN,generated} \ge \mathrm{Lethal\:dose\:of\:HCN}\), then a lethal dose of HCN will be generated in the room when the rug burns.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry, dealing with the quantitative relationships between reactants and products in a chemical reaction. It helps us understand how much of each reactant is needed to form a specific amount of product. To grasp stoichiometry, it's essential to know:

• The balanced chemical equation, which provides the mole ratio of reactants and products.
• The molar masses of each substance involved, allowing for conversion between mass and moles.

In the context of hydrogen cyanide (HCN) calculations, stoichiometry enables us to determine how much sodium cyanide (NaCN) is needed to produce a certain amount of HCN. Based on the equation provided:\(2 \text{NaCN} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HCN}\),we can see that two moles of NaCN react to produce two moles of HCN. This one-to-one mole ratio simplifies calculations, allowing us to directly equate the moles of HCN needed to the moles of NaCN required. Understanding stoichiometry is crucial for safe chemical handling, especially with toxic substances like HCN.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. The specific reaction in question is between sodium cyanide (NaCN) and sulfuric acid (\(\text{H}_2\text{SO}_4\)), which produces hydrogen cyanide gas (HCN). The balanced equation:\[ 2 \text{NaCN} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HCN} \]helps in predicting the amounts of substances consumed and produced during the reaction.
This equation shows that two moles of NaCN react with one mole of \(\text{H}_2\text{SO}_4\) to produce two moles of HCN and one mole of \(\text{Na}_2\text{SO}_4\). Knowing these proportions allows chemists to calculate how much of each reactant is required for a desired amount of product, a crucial part of laboratory and industrial chemistry. Moreover, understanding this reaction is vital for managing materials that produce hazardous substances like HCN. Mastering chemical reactions helps ensure both effective chemical synthesis and safety in environments where these reactions are carried out.

Gas Density

Gas density is an important property that relates the mass of a gas to its volume. It is crucial in calculations involving gases, like when determining the mass of air in a given volume. The density of air at \(26^{\circ} \text{C}\) is \(0.00118 \text{ g/cm}^3\), which can be used to find the mass of air in a room.

• First, calculate the room's volume in cubic meters by multiplying its dimensions. Then, convert this volume to cubic centimeters, as the density of air is given in \(\text{g/cm}^3\).
• Use the formula: \( m_{\text{air}} = V \times \rho_{\text{air}} \), where \( V \) is the volume in \(\text{cm}^3\), and \( \rho_{\text{air}} \) is the density.

This calculation provides the total mass of air present, which can be used to determine the potential distribution of a poisonous gas like HCN within a space. Gas density becomes a particularly critical parameter in safety assessments where the concentration of toxic gases must remain under lethal or hazardous levels.

Empirical Formula

The empirical formula of a compound is the simplest whole-number ratio of the elements within it. For Acrilan ®, used in synthetic fibers, the empirical formula is \(\text{CH}_2\text{CHCN}\). This formula indicates the relative number of each type of atom contained in the fibers.
In the context of HCN production from burning Acrilan ® fibers, understanding empirical formula helps determine how much HCN can be produced. The empirical formula tells us that each unit contains one cyanide group (\(\text{CN}\), which reflects its potential to generate HCN. By calculating the mass portion of HCN in the compound, we found that it constitutes 50.9% of the total mass.

• This percentage helps us assess how much HCN will be produced from the fibers when they burn.
• Even small errors in interpreting these percentages or the empirical formula can vastly affect the predicted safety and risks involved.

Mastering the concept of empirical formulas is essential for chemists working with synthesis and combustion of materials, especially those dealing with toxic substances.