The reaction of potassium superoxide, \(\mathrm{KO}_{2}\), is used in life- support systems to replace \(\mathrm{CO}_{2}(\mathrm{g})\) in expired air with \(\mathrm{O}_{2}(\mathrm{g}) .\) The unbalanced chemical equation for the reaction is given below. $$\mathrm{KO}_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})$$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) are produced by the reaction of \(156 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) with excess \(\mathrm{KO}_{2}(\mathrm{s}) ?\) (b) How many grams of \(\mathrm{KO}_{2}(\mathrm{s})\) are consumed per \(100.0 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) removed from expired air? (c) How many \(\mathrm{O}_{2}\) molecules are produced per milligram of \(\mathrm{KO}_{2}\) consumed?

Step by step solution

01

Balance the chemical reaction

By simply examining the molecule we can balance it as: \(4\ KO_{2}(s)+ 2\ CO_{2}(g) \longrightarrow 2\ K_{2}CO_{3}(s)+ 3\ O_{2}(g)\). This means for every 2 moles of \(CO_{2}(g)\) and 4 moles of \(KO_{2}(s)\) reacted, 3 mol of \(O_{2}(g)\) and 2 moles of \(K_{2}CO_{3}(s)\) are produced.

02

Calculate mol of \(CO_{2}(g)\) from mass and molar mass

First, calculate mol of \(CO_{2}(g)\) reacted using the formula: Mol = Mass/Molar mass. \nGiven mass of \(CO_{2}(g)\) is 156 g, Molar mass of \(CO_{2}(g)\) is 44.01 g/mol. \nTherefore, number of moles = 156 g / 44.01 g/mol = 3.545 mol.

03

Identify mol of \(O_{2}(g)\) produced from molar ratio

Considering the balanced equation, the molar ratio of \(CO_{2}(g)\) to \(O_{2}(g)\) is 2:3, which means for every 2 mol of \(CO_{2}(g)\) reacted, 3 mol of \(O_{2}(g)\) are produced. So, if 3.545 mol of \(CO_{2}(g)\) are reacted, \((3.545 \ mol \ CO_{2} \times 3 \ mol \ O_{2})/2 \ mol \ CO_{2} = 5.318 \ mol \ O_{2}\) will be produced.

04

Calculate mass of \(KO_2(s)\) from mol \(CO_{2}(g)\) and molar ratio

The balanced equation shows that for every 2 mol of \(CO_{2}\) reacted, 4 mol of \(KO_2\) are consumed. Therefore, for 100 g of \(CO_{2}\) which is \(100 \ g \ CO_{2} / 44.01 \ g/mol \ CO_{2} = 2.27 \ mol \ CO_{2}\), we require \(2.27 \ mol \ CO_{2} \times 4 \ mol \ KO_{2} / 2 \ mol \ CO_{2} = 4.54 \ mol \ KO_{2}\). Molar mass of \(KO_2\) is 71.1 g/mol. Hence, the mass of \(KO_{2}\) required is \( 4.54 \ mol \ KO_{2} \times 71.1 \ g/mol \ KO_{2} = 322.84 g \ KO_{2}\).

05

Identify mol of \(O_2\) produced from mol \(KO_2\)

Since the number of molecules is directly proportional to the number of mol, use the molar ratio of \(KO_{2}(s)\) to \(O_{2}(g)\), which is 4:3. For every milligram of \(KO_{2}\) we can find out the number of moles of \(O_{2}\). \n\(1 \ mg \ KO_{2} = 1 \times 10^{-3} g \ KO_{2} = 1 \times 10^{-3} g \ KO_{2} / 71.1 g/mol \ KO_{2} = 1.41 \times 10^{-5} \ mol \ KO_{2}\) \n\(1.41 \times 10^{-5} \ mol \ KO_2 \times 3 \ mol \ O_{2} / 4 \ mol \ KO_{2} = 1.0575 \times 10^{-5} \ mol \ O_{2}\) \nThe number of \(O_{2}\) molecules produced is given by \(1.0575 \times 10^{-5} \ mol \ O_{2} \times Avogadro's \ number (6.02 \times 10^{23} \ molecules/mol) = 6.37 \times 10^{18} \ O_{2} \ molecules \)

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