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Chapter 4: Problem 10

Ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) burns in oxygen. (a) What are the products of the reaction? (b) Write the balanced equation for the reaction. (c) What mass of \(\mathrm{O}_{2}\), in grams, is required for complete combustion of 13.6 of ethane? (d) What is the total mass of products expected from the combustion of \(13.6 \mathrm{g}\) of ethane?

Short Answer

Expert verified
(a) Carbon dioxide and water; (b) \(2\mathrm{C}_2\mathrm{H}_6 + 7\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}\); (c) 50.6 g; (d) 64.2 g.

Step by step solution

01

Identifying Products of Combustion

Ethane (\( \mathrm{C}_2\mathrm{H}_6 \)) burns in oxygen (\( \mathrm{O}_2 \)) generating carbon dioxide (\( \mathrm{CO}_2 \)) and water (\( \mathrm{H}_2\mathrm{O} \)) as products. This is a typical combustion reaction of a hydrocarbon.
02

Balancing the Chemical Equation

Write the unbalanced equation: \[ \mathrm{C}_2\mathrm{H}_6 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]Balance carbon (\( \mathrm{C} \)) atoms by placing 2 \( \mathrm{CO}_2 \): \[ \mathrm{C}_2\mathrm{H}_6 + \mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]Balance hydrogen (\( \mathrm{H} \)) atoms by placing 3 \( \mathrm{H}_2\mathrm{O} \): \[ \mathrm{C}_2\mathrm{H}_6 + \mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O} \]Lastly, balance oxygen (\( \mathrm{O} \)) atoms by placing 7/2 \( \mathrm{O}_2 \) :\[ \mathrm{C}_2\mathrm{H}_6 + \frac{7}{2}\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O} \]Multiply through by 2 to clear the fractional coefficients: \[ 2\mathrm{C}_2\mathrm{H}_6 + 7\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} \]
03

Calculating Mass of Oxygen Required

Use the balanced equation: \[ 2\mathrm{C}_2\mathrm{H}_6 + 7\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} \]Molar mass of ethane: \( 30.08 \, \text{g/mol} \, \mathrm{C}_2\mathrm{H}_6} \)Determine the moles of ethane: \( \frac{13.6 \, \text{g}}{30.08 \, \text{g/mol}} \approx 0.452 } \) molesFrom the balanced equation, 2 moles of \( \mathrm{C}_2\mathrm{H}_6} \) require 7 moles of \( \mathrm{O}_2} \), so 0.452 moles require: \( 0.452 \times \frac{7}{2} \approx 1.582 } \) moles \( \mathrm{O}_2} \)Molar mass of \( \mathrm{O}_2} \): \( 32.00 \, \text{g/mol} \)Mass of \( \mathrm{O}_2} \) needed: \( 1.582 \times 32.00 \approx 50.6 \, \text{g} \)
04

Calculating Total Mass of Products

Using the law of conservation of mass, total mass of reactants = total mass of products.Mass of \( \mathrm{C}_2\mathrm{H}_6} \) = 13.6 \, \text{g}Mass of \( \mathrm{O}_2} \) = 50.6 \, \text{g}Total mass of reactants = 13.6 + 50.6 = 64.2 \, \text{g}Therefore, total mass of products = 64.2 \, \text{g}

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Reactions
Combustion reactions are a type of chemical reaction that happen when a substance combines with oxygen, releasing energy in the form of heat or light. These reactions are particularly common with hydrocarbons, which are compounds made of hydrogen and carbon. When hydrocarbons like ethane (\(\mathrm{C}_2\mathrm{H}_6\)) burn in air, they react with oxygen (\(\mathrm{O}_2\)) to form carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). This process not only produces light and heat, but also results in the complete transformation of the original compound. Understanding combustion reactions is essential because they are a crucial part of our daily lives, from powering vehicles to heating our homes.
Balancing Chemical Equations
Balancing chemical equations is like solving a puzzle where you have to make sure there are equal numbers of each type of atom on both sides of the equation. It's necessary because it reflects the law of conservation of mass, telling us that matter cannot be created or destroyed in a reaction.Here's how you balance a chemical equation step-by-step:
  • Start with the unbalanced equation. For ethane combustion: \[ \mathrm{C}_2\mathrm{H}_6 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]
  • Balance carbon by adding coefficients. Since there are 2 carbons in \(\mathrm{C}_2\mathrm{H}_6\), put a '2' in front of \(\mathrm{CO}_2\).
  • Next, balance hydrogen. \(\mathrm{C}_2\mathrm{H}_6\) has 6 hydrogens, so you need 3 water molecules (\(\mathrm{H}_2\mathrm{O}\)).
  • Finally, balance oxygen by determining how many \(\mathrm{O}_2\) molecules are needed. Oxygen can be tricky, requiring trial and error until each side has equal elements.
  • Clear any fractions by multiplying the entire equation if necessary.
Balancing ensures that all the atoms involved in the reaction are accounted for, respecting the laws of chemistry.
Mole Calculations
Mole calculations are crucial in chemistry for transitioning between mass and number of particles in a chemical substance. When you know the mass of a substance and its molar mass, you can calculate the number of moles, which tells you the actual amount of substance present.For the ethane problem:
  • First, understand the molar mass of ethane: 30.08 g/mol.
  • Then, determine how many moles you have by dividing the mass of ethane by its molar mass: \( \frac{13.6 \, \text{g}}{30.08 \, \text{g/mol}} \approx 0.452 \) moles.
  • Use the balanced chemical equation to find out how many moles of \(\mathrm{O}_2\) are needed. With 0.452 moles of ethane, you need about 1.582 moles of \(\mathrm{O}_2\).
  • Lastly, convert moles of \(\mathrm{O}_2\) into grams using its molar mass: \(32.00 \, \text{g/mol} \).
Mole calculations allow you to transition from theoretical chemistry to practical usage involving real mass measurements.
Conservation of Mass
The law of conservation of mass is foundational in chemistry, asserting that mass cannot be created or destroyed in a chemical reaction. This principle implies that the total mass of the reactants in a chemical reaction is always equal to the total mass of the products. In the combustion of ethane:
  • The mass of ethane (\(\mathrm{C}_2\mathrm{H}_6\)) is 13.6 grams.
  • Calculating the \(\mathrm{O}_2\) needed, you find 50.6 grams.
  • Therefore, the total mass of reactants is 13.6 + 50.6 = 64.2 grams.
  • As per the conservation of mass, the total mass of products should also be 64.2 grams, confirming the principle.
Conservation of mass is key when it comes to balancing chemical equations and performing mole calculations, ensuring that all transformations abide by this foundational law.

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Most popular questions from this chapter

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