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 9.1 g of oxygen in 31 minutes of normal breathing.
express your answer in grams to two significant figures.
mass =

Explanation:

Step1: Calculate the moles of \(O_2\)

The mass of \(O_2\) consumed is \(m = 9.1\space g\). The molar mass of \(O_2\) is \(M = 32\space g/mol\). Using the formula \(n=\frac{m}{M}\), we have \(n_{O_2}=\frac{9.1\space g}{32\space g/mol}\approx0.284\space mol\)

Step2: Use the stoichiometry of the reaction

The balanced chemical equation is \(4KO_2(s)+2CO_2(g)\to2K_2CO_3(s)+3O_2(g)\). From the equation, the mole ratio of \(KO_2\) to \(O_2\) is \(n_{KO_2}:n_{O_2}=4:3\). So \(n_{KO_2}=\frac{4}{3}n_{O_2}\). Substituting \(n_{O_2} = 0.284\space mol\), we get \(n_{KO_2}=\frac{4}{3}\times0.284\space mol\approx0.379\space mol\)

Step3: Calculate the mass of \(KO_2\)

The molar mass of \(KO_2\) is \(M = 71.1\space g/mol\). Using the formula \(m = nM\), we have \(m_{KO_2}=0.379\space mol\times71.1\space g/mol\approx27\space g\)

Answer:

\(27\space g\)