Question

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?

Short answer

(a) 5.318 mol of \(O_{2}(g)\) are produced. \n(b) 322.84 g of \(KO_{2}(s)\) are consumed. \n(c) \(6.37 \times 10^{18}\) \(O_{2}\) molecules are produced.

Step-by-step solution

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.

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.

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.

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 \( 4.54 \ mol \ KO_{2} \times 71.1 \ g/mol \ KO_{2} = 322.84 g \ KO_{2}\).

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 \(1.0575 \times 10^{-5} \ mol \ O_{2} \times Avogadro's \ number (6.02 \times 10^{23} \ molecules/mol) = 6.37 \times 10^{18} \ O_{2} \ molecules \)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in chemistry that ensures the law of conservation of mass is upheld. In a balanced chemical equation, the number of atoms for each element is the same on both sides of the equation. This means that the mass and the number of atoms are conserved throughout the reaction.
To balance a chemical equation, you may follow these simple steps:

• Write down the unbalanced equation.
• List all elements involved in the reaction and count the number of atoms for each.
• Adjust the coefficients (the numbers in front of compounds or elements) to have the same number of atoms of each element on both sides.
• Re-check each element to ensure balance.

In the exercise provided, the initial unbalanced equation is:\[\mathrm{KO}_{2}\mathrm{(s)} + \mathrm{CO}_{2}\mathrm{(g)} \rightarrow \mathrm{K}_{2}\mathrm{CO}_{3}\mathrm{(s)} + \mathrm{O}_{2}\mathrm{(g)}\]Upon balancing, it becomes:\[4\ \mathrm{KO}_{2}\mathrm{(s)} + 2\ \mathrm{CO}_{2}\mathrm{(g)} \rightarrow 2\ \mathrm{K}_{2}\mathrm{CO}_{3}\mathrm{(s)} + 3\ \mathrm{O}_{2}\mathrm{(g)}\]This balanced equation helps in correctly determining the stoichiometric relationships among the reactants and products.

Stoichiometry

Stoichiometry is the quantitative aspect of chemistry that deals with the calculation of reactants and products in chemical reactions. It relies heavily on the concept of the mole and molar ratios from balanced equations.
To perform stoichiometric calculations:

• Ensure the chemical equation is balanced.
• Convert known masses to moles using molar masses.
• Use the molar ratio from the balanced equation to find the moles of the desired substance.
• Convert moles back to mass if needed.

In the exercise, the balanced equation shows that:For every 2 moles of \(CO_{2}(g)\), 3 moles of \(O_{2}(g)\) are produced. When 156 g of \(CO_{2}\) (which is 3.545 moles) react, we calculate the moles of \(O_{2}\) produced as:\[(3.545 \ mol \ \mathrm{CO}_2) \times \left(\frac{3 \ mol \ \mathrm{O}_2}{2 \ mol \ \mathrm{CO}_2}\right) = 5.318 \ mol \ \mathrm{O}_2\]Stoichiometry ensures that you can predict the quantities of reactants needed or the products formed in any chemical reaction.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It serves as a connection between the mass of substances and the amount in moles, crucial for quantitative chemistry calculations.
To calculate the molar mass of a compound:

• Find the atomic masses of each element involved (usually found on the periodic table).
• Multiply the atomic mass of each element by the number of times the element appears in the formula.
• Add these values together to get the total molar mass.

In this exercise:

• The molar mass of \(CO_{2}\) is calculated as: C (12.01 g/mol) + O (16.00 g/mol)×2 = 44.01 g/mol.
• For \(KO_2\), calculate as: K (39.10 g/mol) + O (16.00 g/mol)×2 = 71.1 g/mol.

Understanding molar mass allows for the conversion between mass and moles of any substance, facilitating stoichiometric calculations.

Avogadro's Number

Avogadro's Number is a key concept in chemistry, defined as the number of atoms, ions, or molecules in one mole of a substance. This number is approximately \(6.02 \times 10^{23}\). It provides the link between the macroscopic scale of material quantities and the microscopic scale of individual atoms and molecules.
Avogadro's Number lets chemists understand how many particles they are dealing with when measuring moles. So if one mole contains \(6.02 \times 10^{23}\) entities, then understanding this concept helps in converting between moles and molecular count.
In the exercise, we calculate the number of \(O_{2}\) molecules produced per milligram of \(KO_{2}\) as:Number of \(O_{2}\) molecules = Moles of \(O_{2}\) × Avogadro's NumberFor instance, if 0.001 g of \(KO_{2}\) produces \(1.0575 \times 10^{-5}\) mol of \(O_{2}\), we calculate:Number of \(O_{2}\) molecules = \(1.0575 \times 10^{-5}\ mol \ \times 6.02 \times 10^{23}\ mol^{-1} = 6.37 \times 10^{18}\) O\(_2\) molecules.This shows the beauty of Avogadro's Number in relating the amount of a substance to the number of constituent particles.