   # A mixture of butene, C 4 H g , and butane, is burned in air to give CO 2 and water. Suppose you burn 2.86 g of the mixture and obtain 8.80 g of CO 2 and 4.14 g of H 2 O. What are the mass percentages of butene and butane in the mixture? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 101GQ
Textbook Problem
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## A mixture of butene, C4Hg, and butane, is burned in air to give CO2 and water. Suppose you burn 2.86 g of the mixture and obtain 8.80 g of CO2 and 4.14 g of H2O. What are the mass percentages of butene and butane in the mixture?

Interpretation Introduction

Interpretation:

The mass percent of butane and butene in the given mixture has be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is,

Numberofmole=Mass in gramMolar mass

Mass in gramofthesubstance=Numberofmole×Molarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.

### Explanation of Solution

Here 8.80gofCO2and4.14gofH2O is obtained after heating 2.86g mixture of butane and butene.

The balanced equation for the reaction of burning a mixture of butane and butene is,

C4H8+C4H108CO2+9H2O

The amount of CO2 =8.80gCO244.01g/mol = 0.199molCO2

The amount of H2O  = 4.14gH2O18.0g/molH2O =0.23 mol H2O

In accordance with the amount of product produced in the reaction it is clear that carbon dioxide acts as limiting reagent.

mass of butene  = 0.199molCO2×1butene8CO2×56

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