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Chapter 3: Problem 111

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 \(12 \times 15 \times 8.0 \mathrm{ft}\). The density of air at \(26^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3}\). (b) If the \(\mathrm{HCN}\) is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what mass of \(\mathrm{NaCN}\) 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 \(^{\oplus}\) or Acrilan \(^{\otimes}\) burn. Acrilan \(^{\oplus}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so \(\mathrm{HCN}\) is \(50.9 \%\) of the formula by mass. \(\mathrm{A}\) rug measures \(12 \times 15 \mathrm{ft}\) and contains 30 oz of Acrilan \(^{\otimes}\) 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
The lethal dose of Hydrogen Cyanide (HCN) in the small laboratory room is calculated to be \(m_\mathrm{HCN}\). To produce this lethal dose, a mass of NaCN is required, which is calculated using the balanced chemical equation. Burning the Acrilan rug generates an amount of HCN, considering the yield and consumption. Comparing this generated HCN to the lethal dose determines if the burning of the rug could produce a lethal dose of HCN in the room.

Step by step solution

01

Calculate the volume of the laboratory room

To calculate the volume, use the given dimensions: \(12 \times 15 \times 8.0 \mathrm{ft}\). Convert these dimensions from feet to centimeters. Note that there are 30.48 centimeters in one foot. \[V_\text{room} = 12\text{ ft} \times 15\text{ ft} \times 8.0\text{ ft} \times \left(\frac{30.48\text{ cm}}{1\text{ ft}}\right)^3\]
02

Calculate the mass of air in the room

To find the mass of air in the room, use the volume of the room and the given density of air at \(26^{\circ} \mathrm{C}\). \[m_\text{air} = V_\text{room} \times 0.00118 \frac{\text{g}}{\text{cm}^3}\]
03

Calculate the lethal dose of HCN in the room

To find the lethal dose of HCN, multiply the mass of air by the lethal dose per kilogram of air: \[m_\mathrm{HCN} = m_\text{air} \times \frac{300\text{ mg HCN}}{1\text{ kg air}}\]
04

Calculate the mass of NaCN required to form the lethal dose of HCN

Using the balanced chemical equation, find the molar ratio between NaCN and HCN: \[2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g)\] Using this ratio, calculate the mass of NaCN required to form the lethal dose of HCN. Remember to convert HCN mass to moles and use the molar mass of NaCN to find its mass.
05

Calculate the amount of HCN generated by burning the rug

First, find the total mass of Acrilan fibers in the rug, using the fiber amount per square yard and the rug area. Next, find the mass of HCN in the Acrilan fibers using the given percentage composition of HCN (50.9%). Finally, calculate the amount of HCN generated by burning the rug by considering the given conditions: assume 20% HCN yield and 50% rug consumption.
06

Compare the generated HCN with the lethal dose

The final step is to compare the calculated amount of HCN generated by burning the rug to the lethal dose calculated in step 3. If the amount of generated HCN is equal to or greater than the lethal dose, then a lethal dose will be produced in the room when the rug burns.

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Most popular questions from this chapter

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.0130 \mathrm{~mol} \mathrm{C}, 0.0390 \mathrm{~mol} \mathrm{H},\) and \(0.0065 \mathrm{~mol} \mathrm{O} ;\) (b) \(11.66 \mathrm{~g}\) iron and \(5.01 \mathrm{~g}\) oxygen; (c) \(40.0 \% \mathrm{C}, 6.7 \% \mathrm{H},\) and \(53.3 \% \mathrm{O}\) by mass.

Detonation of nitroglycerin proceeds as follows: $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}_{3} \mathrm{O}_{9}(l) \longrightarrow \\ \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad12 \mathrm{CO}_{2}(g)+6 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) If a sample containing 2.00 \(\mathrm{mL}\) of nitroglycerin (density \(=\) 1.592 \(\mathrm{g} / \mathrm{mL}\) ) is detonated, how many moles of gas are produced? (b) If each mole of gas occupies 55 Lunder the conditions of the explosion, how many liters of gas are produced? (c) How many grams of \(\mathrm{N}_{2}\) are produced in the detonation?

When a mixture of \(10.0 \mathrm{~g}\) of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right) ;\) and \(10.0 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resultant combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

(a) What is the mass, in grams, of \(2.50 \times 10^{-3} \mathrm{~mol}\) of ammonium phosphate? (b) How many moles of chloride ions are in \(0.2550 \mathrm{~g}\) of aluminum chloride? (c) What is the mass, in grams, of \(7.70 \times 10^{20}\) molecules of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\) (d) What is the molar mass of cholesterol if \(0.00105 \mathrm{~mol}\) has a mass of \(0.406 \mathrm{~g} ?\)

What parts of balanced chemical equations give information about the relative numbers of moles of reactants and products involved in a reaction?
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