1) Écrire les équations stœchiométriques pour la combustion...

1) Écrire les équations stœchiométriques pour la combustion du méthane, $\mathrm{CH}_4$ et de l'éthane, $\mathrm{C}_2 \mathrm{H}_4$. 2) La combustion de $1,210 \mathrm{~g}$ d'un mélange de méthane et d'éthane a fourni $3,372 \mathrm{~g}$ de $\mathrm{CO}_2$. Déduire de ce résultat la composition du mélange exprimée en fraction molaire puis en fraction massique.

1) Écrire les équations stœchiométriques pour la combustion du méthane, $\mathrm{CH}_4$ et de l'éthane, $\mathrm{C}_2 \mathrm{H}_4$.
2) La combustion de $1,210 \mathrm{~g}$ d'un mélange de méthane et d'éthane a fourni $3,372 \mathrm{~g}$ de $\mathrm{CO}_2$. Déduire de ce résultat la composition du mélange exprimée en fraction molaire puis en fraction massique.

Key Concepts

Balanced Chemical Equations

Balanced chemical equations are a fundamental representation of chemical reactions that ensure the conservation of atoms. They involve assigning stoichiometric coefficients so that the number of atoms for each element is the same on both sides of the equation, which is essential for accurately relating the quantities of reactants and products.

Combustion Reactions of Hydrocarbons

Combustion reactions, particularly of hydrocarbons, are processes where a hydrocarbon reacts with oxygen to form carbon dioxide and water. This type of reaction is exothermic and the balanced equations derived from it are crucial for determining the molar ratios involved, which are necessary for subsequent quantitative analysis.

Stoichiometric Analysis

Stoichiometric analysis involves using the balanced chemical equations to relate the moles and masses of reactants and products. This process provides the tools for converting between mass and mole quantities, allowing for the calculation of the amounts of substances consumed or produced in a reaction.

Mole and Mass Fractions

Mole fraction is defined as the ratio of the number of moles of a component to the total number of moles in a mixture, whereas mass fraction is the ratio of the mass of a component to the total mass of the mixture. Both are key concepts in describing the composition of mixtures, enabling the expression of component distributions in different terms.

Transcript

00:01 In this reaction, we have a mixture of hydrocarbons reacting with oxygen in a combustion reaction to form carbon dioxide and water.

00:10 To find the mass of each of the two hydrocarbons, ch4 and c2h6, in the mixture, we need to look at the mole ratio between carbon dioxide and these compounds.

00:24 For every one mole of ch4, this relates to one mole of co2.

00:37 Because there's one carbon in each.

00:40 If, however, i have one mole of c2h6, this relates to two moles of carbon dioxide because it's a one to two ratio.

00:55 And so if i can find the moles of co2, this is equal to my moles of ch4 plus the moles of c2 -8.

01:16 Times a 2 to 1 ratio.

01:27 The moles of co2 are equal to the grams of co2 divided by the molar mass, or 0 .04691 moles of co2.

01:59 The moles of ch4 would be equal to the mass of ch4, divided by the molar mass of ch4, which is 16.

02:23 And the moles of c2h6, similarly, is equal to the mass of c2h6, divided by its molar mass, which is 30 grams per mole.

02:45 So if we let x equal our mass of ch4, my moles of ch4 is equal to x over 16.

03:03 And if x is the mass of ch4, then my total mass, 0 .732 minus x is equal to the mass of c2h6.

03:17 So my moles, c2h6, is equal to 0 .732 minus x divided by 6.

03:29 30.

03:31 And now we can substitute these into our equation above here...

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