Question

oxygen masks for producing o₂ in emergency situations contain potassium superoxide (ko₂). it reacts according to the following equation: 4ko₂ + 2h₂o + 4co₂ → 4khco₃ + 3o₂ (a) if a person wearing such a mask exhales 0.85 g of co₂ every minute, how many moles of ko₂ are consumed in 10.0 minutes? mol ko₂ etextbook and media (b) how many grams of oxygen are produced in 1.0 hour? g o₂

Explanation:

Step1: Calculate the mass of \(CO_2\) exhaled in 10.0 minutes

The mass of \(CO_2\) exhaled per minute is \(0.85\ g\). So, in \(10.0\) minutes, the mass of \(CO_2\), \(m(CO_2)=0.85\ g/min\times10.0\ min = 8.5\ g\)

Step2: Calculate the moles of \(CO_2\)

The molar mass of \(CO_2\), \(M(CO_2)=(12.01 + 2\times16.00)\ g/mol=44.01\ g/mol\)
The moles of \(CO_2\), \(n(CO_2)=\frac{m(CO_2)}{M(CO_2)}=\frac{8.5\ g}{44.01\ g/mol}\approx0.193\ mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(4KO_2 + 2H_2O+4CO_2
ightarrow4KHCO_3 + 3O_2\), the mole ratio of \(KO_2\) to \(CO_2\) is \(4:4 = 1:1\)
So, the moles of \(KO_2\) consumed, \(n(KO_2)=n(CO_2)\approx0.193\ mol\)

Step1: Calculate the mass of \(CO_2\) exhaled in 1.0 hour

Since \(1\) hour \( = 60\) minutes, and the mass of \(CO_2\) exhaled per minute is \(0.85\ g\), the mass of \(CO_2\), \(m(CO_2)=0.85\ g/min\times60\ min = 51\ g\)

Step2: Calculate the moles of \(CO_2\)

The molar mass of \(CO_2\), \(M(CO_2) = 44.01\ g/mol\)
The moles of \(CO_2\), \(n(CO_2)=\frac{m(CO_2)}{M(CO_2)}=\frac{51\ g}{44.01\ g/mol}\approx1.16\ mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(4KO_2 + 2H_2O + 4CO_2
ightarrow4KHCO_3+3O_2\), the mole ratio of \(O_2\) to \(CO_2\) is \(3:4\)
The moles of \(O_2\) produced, \(n(O_2)=\frac{3}{4}n(CO_2)=\frac{3}{4}\times1.16\ mol = 0.87\ mol\)

Step4: Calculate the mass of \(O_2\)

The molar mass of \(O_2\), \(M(O_2)=32.00\ g/mol\)
The mass of \(O_2\), \(m(O_2)=n(O_2)\times M(O_2)=0.87\ mol\times32.00\ g/mol = 27.84\ g\)

Answer:

\(0.193\)

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