/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 113 Hydrogen cyanide, HCN, is a pois... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mathrm{HCN}\) is formed by reaction of 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 \(^{\circledast}\) 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

Expert verified
(a) 13.94 g of HCN is the lethal dose in the room. (b) 25.27 g of NaCN generates the lethal dose. (c) Yes, burning the rug generates more HCN than this lethal dose.

Step by step solution

01

Calculate Room Volume

First, we need to calculate the volume of the room in cubic meters: \[ \text{Volume} = 3.5 \times 4.5 \times 2.5 = 39.375 \text{ m}^3 \]
02

Convert Air Density to kg/m³

The density of air is given in \( \text{g/cm}^3 \). Convert it to \( \text{kg/m}^3 \): \[ 0.00118 \text{ g/cm}^3 = 1.18 \text{ kg/m}^3 \]
03

Calculate Mass of Air in the Room

Use the room volume and the air density to find the mass of air: \[ \text{Mass of air} = 1.18 \text{ kg/m}^3 \times 39.375 \text{ m}^3 = 46.4625 \text{ kg} \]
04

Determine Lethal Dose of HCN

The lethal dose of HCN is given as \( 300 \text{ mg/kg} \). Convert to grams and calculate for the entire room: \[ \text{Lethal Dose} = 300 \text{ mg/kg} \times 46.4625 \text{ kg} = 13938.75 \text{ mg} = 13.93875 \text{ g} \]
05

Calculate Mass of NaCN Needed

From the chemical equation, \( 2 \text{ mol NaCN} \) produces \( 2 \text{ mol HCN} \). Molar mass of HCN is about 27.03 g/mol. To produce 13.93875 g HCN, we need:\[ \text{Moles of HCN} = \frac{13.93875 \text{ g}}{27.03 \text{ g/mol}} = 0.5158 \text{ mol} \]Therefore, 0.5158 mol NaCN is needed. With a molar mass of about 49.01 g/mol:\[ \text{Mass of NaCN} = 0.5158 \text{ mol} \times 49.01 \text{ g/mol} = 25.27 \text{ g} \]
06

Calculate if Rug Burns Release Lethal Dose

Convert room dimensions to yards (1 meter = 1.094 yards) to find carpet area:\[ 3.5 \times 1.094 \times 4.5 \times 1.094 = 18.54 \text{ yd}^2 \]Carpet weight is \( 850 \text{ g/yd}^2 \), so total weight is \( 18.54 \times 850 = 15759 \text{ g} \).Since Acrilan is 50.9% HCN by mass:\[ 15759 \text{ g} \times 0.509 \times 0.2 \times 0.5 = 1599.97 \text{ g HCN} \]The amount of HCN produced (1599.97 g) exceeds the lethal dose (13.93875 g).
07

Conclusion

If the carpet burns, it will produce much more HCN than the lethal dose needed for the room, potentially making the environment lethal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemical reactions are processes where substances change into new substances with different properties. In the equation provided, sodium cyanide (aCN) reacts with sulfuric acid (a_2SO_4) to form sodium sulfate and hydrogen cyanide (HCN): \[2 \text{NaCN} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HCN}\] In this reaction, two molecules of sodium cyanide produce two molecules of hydrogen cyanide. **Key Points of Chemical Reactions:**
  • Reactants are substances that start the reaction (here, NaCN and \( H_2SO_4 \)).
  • Products are the new substances formed (here, \( Na_2SO_4 \) and \( HCN \)).
  • Stoichiometry reveals the proportions of each substance used and produced based on the balanced equation.
For safety, it is crucial to understand these reactions, especially when hazardous substances like HCN are involved.
Molar Mass Calculations
Molar mass calculations are essential for determining how much of a substance is involved in a chemical reaction. Molar mass is the mass of one mole of a substance. **Calculating Molar Mass:**
  • To find the molar mass of a compound, sum the atomic masses of all atoms in the formula. Oxygen has an atomic mass of about 16, so O in HCN contributes 16 units, for example.
  • For hydrogen cyanide, \( HCN \): \[ \text{Molar mass} = \text{H} + \text{C} + \text{N} = 1.01 + 12.01 + 14.01 = 27.03 \, \text{g/mol} \]
  • The molar mass of \( NaCN \) is about 49.01 g/mol, combining sodium (Na), carbon (C), and nitrogen (N).
These calculations help in determining how much of a reactant (e.g., \( NaCN \)) is needed to produce a specific amount of product (e.g., \( HCN \)).
Density Conversion
Density is a measure of how much mass is contained in a specific volume. When dealing with gases like air, density is often needed in different units to calculate mass from volume. **Converting Density Units:**
  • The given density of air is \(0.00118 \text{ g/cm}^3\).
  • When converting to \( \text{kg/m}^3\): - Multiply by 1000 (to move from grams to kilograms). - Multiply by 1,000,000 because \(1 \text{ m}^3 = 1,000,000 \text{ cm}^3\).
  • The converted density is: \[0.00118 \text{ g/cm}^3 \times 1000 \times 1,000,000 = 1.18 \text{ kg/m}^3\]
Understanding density conversion helps accurately calculate how much of a toxic gas could fill a space and pose a danger.

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Most popular questions from this chapter

Fructose, commonly called fruit sugar, is a monosaccharide found in many plants. It contains \(40 \% \mathrm{C}, 6.71 \% \mathrm{H}\), and the remainder O. (a) What is the empirical formula for fructose? (b) A mass spectrum of fructose shows a peak at about \(180 \mathrm{u}\). What is the molecular formula of the substance?

Determine the empirical formula of each of the following compounds if a sample contains (a) \(3.92 \mathrm{~mol} \mathrm{C}, 5.99 \mathrm{~mol} \mathrm{H}\) and \(2.94 \mathrm{~mol} \mathrm{O} ;(\mathbf{b}) 12.0 \mathrm{~g}\) calcium and \(2.8 \mathrm{~g}\) nitrogen; \((\mathbf{c})\) \(89.14 \%\) Au and \(10.86 \%\) O by mass.

When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed if \(1.25 \mathrm{~g}\) of hydrogen sulfide is bubbled into a solution containing \(2.00 \mathrm{~g}\) of sodium hydroxide, assuming that the sodium sulfide is made in \(92.0 \%\) yield?

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\), but other products containing \(\mathrm{Cl}\), such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\), are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\). (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}\) and \(\mathrm{HCl}\). (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

(a) You are given a cube of silver metal that measures 1.000 \(\mathrm{cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

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