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Chapter 8: Problem 118

A mixture of methane and ethylene in the ratio of a:b by volume occupies \(30 \mathrm{ml}\). On complete combustion, the mixture yield \(40 \mathrm{ml}\) of \(\mathrm{CO}_{2} .\) What volume of \(\mathrm{CO}_{2}\) would have been obtained if the ratio would have been b:a? (a) \(50 \mathrm{ml}\) (b) \(30 \mathrm{ml}\) (c) \(40 \mathrm{ml}\) (d) \(60 \mathrm{ml}\)

Short Answer

Expert verified
Answer: 50 mL

Step by step solution

01

Write balanced chemical equations for combustion of methane and ethylene

Firstly, let's write the balanced chemical equations for the combustion of methane and ethylene, respectively. Methane combustion: CH4 + 2O2 -> CO2 + 2H2O Ethylene combustion: C2H4 + 3O2 -> 2CO2 + 2H2O These equations will help us to determine the moles of methane and ethylene in the a:b ratio and also find the amount of CO2 generated.
02

Determine the moles of methane and ethylene in the mixture

Let's say the moles of methane in the mixture are "a". According to the balanced chemical equation, 1 mole of methane produces 1 mole of CO2. Therefore, "a" moles of methane produce "a" moles of CO2. Similarly, we assume that the moles of ethylene present in the mixture are "b". As per the balanced chemical equation, 1 mole of ethylene generates 2 moles of CO2. Thus, "b" moles of ethylene produce "2b" moles of CO2.
03

Calculate the total volume of CO2 produced for the a:b ratio

Given the volume of CO2 produced in the a:b ratio is 40 mL. From the chemical equations, we can write the equation as: a + 2b = 40
04

Calculate the proportionality factor for methane and ethylene

Now, the sum of the moles of methane and ethylene in the mixture is given by their proportions a and b: a + b = 30 Using this equation, we can find the relationship between a and b. Let's substitute a = 30 - b in the previous equation: (30 - b) + 2b = 40 b = 10 Now substituting the value of b in a+b=30, a = 30 - 10 = 20
05

Find the volume of CO2 produced on complete combustion in the b:a ratio

In this case, the ratio of methane and ethylene moles changes from a:b to b:a. Now, we have b moles of methane and a moles of ethylene. Using the balanced chemical equations, we can write the equation for this case as: b + 2a = 10 + 2(20) = 10 + 40 = 50 Thus, the volume of CO2 produced when the ratio of methane and ethylene are in b:a is 50 mL (Option a).

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Most popular questions from this chapter

Mole fraction of solute in an aqueous solution of \(\mathrm{NaOH}\) is \(0.1 .\) If the specific gravity of the solution is \(1.4\), the molarity of the solution is (a) \(6.93\) (b) \(0.1\) (c) \(71.4\) (d) \(0.14\)

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