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3 years ago

Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 12.0 ft×15.0 ft×8.60ft.

Chemistry

1 answer:

garik1379 [7]3 years ago

7 0

Given:

Dimensions of the room= 12 ft * 15 ft * 8.60 ft

To determine:

The amount of HCN that gives the lethal dose in the room with the given dimensions

Explanation:

As per the World Health Organization, the lethal dose of HCN is around 300 ppm

300 ppm = 300 mg of HCN/ kg of inhaled air

Volume of air = volume of room = 12 * 15 *8.6 = 1548 ft³

Now, 1 ft³ = 28316.8 cm³

Therefore, the calculated volume of air corresponds to:

1548 * 28316.8 = 4.383 * 10⁷ cm3

Density of air (at room temperature 25 C) = 0.00118 g/cm3

Thus mass of air corresponding to the calculated volume is

Mass = Density * volume = 0.00118 g/cm3 * 4.383 * 10⁷ cm3

= 5.172*10⁴ g = 51.72 kg

Lethal amount of HCN corresponding to 51.72 kg of air would be.

= 51.72 kg air* 300 mg of HCN/1 kg air = 15516 mg

Ans: Lethal dose of HCN = 15.5 g

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The question is incomplete, here is the complete question:

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The equation used to calculate number of moles is given by:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

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$V_2\text{ and }n_2$ are the volume and number of moles of hydrogen chloride

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$V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol$

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