When an oxygen mask is deployed, the flow rate of oxygen gas is designed to change over time as the airplane falls to a safe altitude. Shown below is the manufacturer's specification for the flow

Question

When an oxygen mask is deployed, the flow rate of oxygen gas is designed to change over time as the airplane falls to a safe altitude. Shown below is the manufacturer's specification for the flow rate from one such oxygen generator.

Ten seconds after being activated, the flow rate is at its maximum of $3.6 d m^{3} mi n^{- 1}$ . This lasts for approximately 50 seconds before falling as shown in the graph. Calculate the amount of KClO₃ required by the oxygen generator for 12 minutes to produce sufficient oxygen if the yield of the reaction is 80%. [Given: 1 mol of the gas occupies 24 L under the reaction temperature and pressure; Atomic masses: K = 39, Cl = 35.5, O = 16]

TestHub · Accepted Answer

The volume of oxygen required is the area under the curve which works out to be 14.975 $dm^{3}$ which is equivalent to 0.62 mol of $O_{2}$ .

$2 KClO_{3} ⟶ 2 KCl + 3 O_{2}$

$W_{KClO_{3}} = \frac{2}{3} \times 0.62 \times \frac{100}{80} \times 122.5 = 63.29 g$

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