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Chapter 8: Problem 22

The mass of \(\mathrm{CO}_{2}\) that must be mixed with \(20 \mathrm{~g}\) of oxygen such that \(27 \mathrm{ml}\) of a sample of the resulting mixture would contain equal number of molecules of each gas (a) \(13.75 \mathrm{~g}\) (b) \(27.50 \mathrm{~g}\) (c) \(41.25 \mathrm{~g}\) (d) \(55 \mathrm{~g}\)

Short Answer

Expert verified
Answer: 27.5 g

Step by step solution

01

Identify the known variables

We are given the following information: - Mass of oxygen (O₂): 20 g - Volume of the mixture: 27 ml Our goal is to find the mass of CO₂ required to meet the condition of having the same number of molecules in 27 ml of the mixture.
02

Calculate the moles of oxygen

We shall start by determining the number of moles of O₂. We can use the formula: Moles of O₂ = mass of O₂/ molar mass of O₂ The molar mass of O₂ is 32 g/mol (since each oxygen atom has a molar mass of 16 g/mol). Given that the mass of O₂ is 20 g, we can calculate the moles of O₂: Moles of O₂ = 20 g / 32 g/mol = 0.625 mol
03

Calculate the moles of CO₂ needed to have the same number of molecules as O₂

Since we need the same number of molecules for both gases in the mixture, we need the same number of moles for CO₂ as moles of O₂. So, the moles of CO₂ should also be 0.625 mol.
04

Calculate the mass of CO₂ required

To find the mass of CO₂ we need, we will use the formula: Mass of CO₂ = moles of CO₂ * molar mass of CO₂ The molar mass of CO₂ is 44 g/mol (since carbon has a molar mass of 12 g/mol and each oxygen atom has a molar mass of 16 g/mol, giving a combined molar mass of 12 g/mol + (2*16 g/mol) = 44 g/mol). Now we can calculate the mass of CO₂ required: Mass of CO₂ = 0.625 mol * 44 g/mol = 27.5 g So the mass of CO₂ that must be mixed with 20 g of oxygen to obtain a mixture in which 27 ml of the sample contains an equal number of molecules of each gas is 27.5 g. Therefore, the correct answer is (b) \(27.50 \mathrm{~g}\).

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Most popular questions from this chapter

A quantity of \(1.4 \mathrm{~g}\) of a hydrocarbon gives \(1.8 \mathrm{~g}\) water on complete combustion. The empirical formula of hydrocarbon is (a) \(\mathrm{CH}\) (b) \(\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}\) (d) \(\overline{\mathrm{CH}}\),

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