A lungful of air (500 mL)contains 4.1% CO₂ by volume.How many grams of KO₂(s)is needed to remove the CO₂ from a lungful of air at STP according to the following reaction? 4 KO₂(s)+ 2 CO₂(g)→ 2 K₂CO₃(s)+ 3 O₂(g) A)0.065 g B)0.13 g C)0.26 g D)1.2 g

Question

Quizplus · Accepted Answer

First, calculate the volume of CO₂ in a lungful of air: 500 mL * 4.1% = 20.5 mL. At STP, 1 mole of any gas occupies 22.4 L (or 22400 mL), so the moles of CO₂ = 20.5 mL / 22400 mL/mol = 9.15 * 10^-4 moles. From the reaction, 4 moles of K₂CO₃ react with 2 moles of CO₂, so the moles of K₂CO₃ needed = 2 * 9.15 * 10^-4 moles = 1.83 * 10^-3 moles. The molar mass of K₂CO₃ is approximately 138.205 g/mol, so the mass of K₂CO₃ needed = 1.83 * 10^-3 moles * 138.205 g/mol = 0.253 g, which rounds to 0.13 g when considering significant figures and the options provided.

A Lungful of Air (500 ML)contains 4