Question

an emergency breathing apparatus often placed in mines or caves works via the following chemical reaction. 4ko₂(s) + 2co₂(g) → 2k₂co₃(s) + 3o₂(g) if the oxygen supply becomes limited or if the air becomes poisoned, a worker can use the apparatus to breathe while exiting the mine. notice that the reaction produces o₂, which can be breathed, and absorbs co₂, a product of respiration. part a what minimum amount of ko₂ is required for the apparatus to produce enough oxygen to allow the user 31 minutes to exit in an emergency? assume that an adult consumes approximately 91 g of oxygen in 31 minutes of normal breathing. express your answer in grams to two significant figures. < return to assignment provide feedback submit request answer mass = 8

Explanation:

Step1: Calculate the amount of \(O_2\) needed

An adult consumes approximately \(91\ g\) of \(O_2\) in \(31\) minutes. So the amount of \(O_2\) needed is \(m_{O_2}=91\ g\).

Step2: Use the stoichiometry of the reaction

The balanced chemical reaction is \(4KO_2(s)+2CO_2(g)\to2K_2CO_3(s)+3O_2(g)\).
From the reaction, the mole ratio of \(KO_2\) to \(O_2\) is \(n_{KO_2}:n_{O_2}=4:3\).
The molar mass of \(O_2\) is \(M_{O_2} = 32\ g/mol\), and the molar mass of \(KO_2\) is \(M_{KO_2}=(39 + 2\times16)=71\ g/mol\).
First, find the number of moles of \(O_2\): \(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{91\ g}{32\ g/mol}\).
Then, find the number of moles of \(KO_2\) using the mole ratio: \(n_{KO_2}=\frac{4}{3}n_{O_2}=\frac{4}{3}\times\frac{91\ g}{32\ g/mol}\).
Finally, find the mass of \(KO_2\): \(m_{KO_2}=n_{KO_2}\times M_{KO_2}=\frac{4}{3}\times\frac{91\ g}{32\ g/mol}\times71\ g/mol\).

$$
LATEXBLOCK0
$$

Answer:

\(2.7\times 10^{2}\ g\)