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Ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\), burns in oxygen to give carbon dioxide, \(\mathrm{CO}_{2}\), and water. Write the equation for the reaction, giving molecular, molar, and mass interpretations below the equation.

Short Answer

Expert verified
The reaction is: \( \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \).

Step by step solution

01

Write the Balanced Chemical Equation

When ethylene (\( \text{C}_2\text{H}_4 \)) burns in oxygen (\( \text{O}_2 \)), it produces carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)). The balanced chemical equation for this reaction is: \[ \text{C}_2\text{H}_4 (g) + 3\text{O}_2 (g) \rightarrow 2\text{CO}_2 (g) + 2\text{H}_2\text{O} (g) \].
02

Molecular Interpretation

In this equation, one molecule of ethylene reacts with three molecules of oxygen to produce two molecules of carbon dioxide and two molecules of water.
03

Molar Interpretation

According to the balanced equation, one mole of ethylene reacts with three moles of oxygen to produce two moles of carbon dioxide and two moles of water. This reflects the stoichiometry of the reaction using moles.
04

Mass Interpretation

Based on the molar mass of each substance, the reaction can also be interpreted in terms of mass. Using molar masses: ethylene (\(28.05 \text{ g/mol} \)), oxygen (\(32.00 \text{ g/mol} \)), carbon dioxide (\(44.01 \text{ g/mol} \)), and water (\(18.02 \text{ g/mol} \)), the mass interpretation corresponds to the equation: \(28.05\text{ g of }\text{C}_2\text{H}_4 + 96.00\text{ g of }\text{O}_2 \rightarrow 88.02\text{ g of }\text{CO}_2 + 36.04\text{ g of }\text{H}_2\text{O} \).

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Key Concepts

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

Molecular Interpretation
In the balanced reaction of ethylene with oxygen, we understand the molecular interpretation by breaking down the interaction at the level of individual molecules. When interpreting the reaction at the molecular level, we say that one molecule of ethylene (\( \text{C}_2\text{H}_4 \)) combines with three molecules of oxygen gas (\( \text{O}_2 \)). This transformation results in the formation of two molecules of carbon dioxide (\( \text{CO}_2 \)) and two molecules of water (\( \text{H}_2\text{O} \)).
This molecular view helps us imagine how each molecule of ethylene interacts with molecules of oxygen. It's akin to relating individual puzzle pieces coming together to form a complete picture.
  • Reactants: 1 molecule of ethylene + 3 molecules of oxygen
  • Products: 2 molecules of carbon dioxide + 2 molecules of water
Understanding chemical reactions at the molecular level is crucial for comprehending how substances change during a chemical reaction, helping you visualize the perfect dance of atoms forming new compounds.
Molar Interpretation
The molar interpretation extends our view from individual molecules to a larger scale, using moles as a measure. A mole is a standard unit in chemistry that denotes \( 6.022 \times 10^{23} \) entities, allowing chemists to work with manageable figures instead of counting atoms or molecules.For the reaction of ethylene with oxygen:
  • 1 mole of ethylene reacts with 3 moles of oxygen.
  • This produces 2 moles of carbon dioxide and 2 moles of water.
This conversion to moles mirrors the stoichiometry seen in the balanced equation and allows calculations regarding how many moles of each substance are involved in or formed by the reaction. It is key for predicting how much product will form or reactants will be needed based on initial mole quantities. With the molar interpretation, chemists can also link physical quantities to chemical formulas, providing a practical way to measure and control reactions in the lab.
Mass Interpretation
In mass interpretation, we take a practical perspective by converting moles into grams using molar masses. This approach uses the actual weights of compounds to predictively track the progress of a reaction or ensure its feasible scalability.
Using molar masses:
  • Ethylene: 28.05 g/mol
  • Oxygen: 32.00 g/mol
  • Carbon dioxide: 44.01 g/mol
  • Water: 18.02 g/mol
Based on this, the balanced equation suggests the following mass equivalencies:
\( 28.05 \text{ g of ethylene} + 96.00 \text{ g of oxygen} \rightarrow 88.02 \text{ g of carbon dioxide} + 36.04 \text{ g of water} \). Mass interpretation is integral for practical laboratory work since it connects measured quantities — equipment like balances readily measure mass — to the stoichiometric ratios of chemical reactions. It simplifies the complex nature of chemical reactions into weighable quantities to aid in efficient and precise reaction planning.

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Most popular questions from this chapter

An oxide of tungsten (symbol \(\mathrm{W}\) ) is a bright yellow solid. If \(5.34 \mathrm{~g}\) of the compound contains \(4.23 \mathrm{~g}\) of tungsten, what is its empirical formula?

A sample of methane gas, \(\mathrm{CH}_{4}(g)\), is reacted with oxygen gas to produce carbon dioxide and water. If \(20.0 \mathrm{~L}\) of methane (density \(=1.82 \mathrm{~kg} / \mathrm{m}^{3}\) ) and \(30.0 \mathrm{~L}\) of oxygen gas (density \(=1.31 \mathrm{~kg} / \mathrm{m}^{3}\) ) are placed into a container and allowed to react, how many \(\mathrm{kg}\) of carbon dioxide will be produced by the reaction?

A power plant is driven by the combustion of a complex fossil fuel having the formula \(\mathrm{C}_{11} \mathrm{H}_{7} \mathrm{~S}\). Assume the air supply is composed of only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) with a molar ratio of 3.76: 1.00 , and the \(\mathrm{N}_{2}\) remains unreacted. In addition to the water produced, the fuel's \(\mathrm{C}\) is completely combusted to \(\mathrm{CO}_{2}\) and its sulfur content is converted to \(\mathrm{SO}_{2}\). In order to evaluate gases emitted at the exhaust stacks for environmental regulation purposes, the nitrogen supplied with the air must also be included in the balanced reactions. Including the \(\mathrm{N}_{2}\) supplied in the air, write a balanced combustion equation for the complex fuel assuming \(100 \%\) stoichiometric combustion (i.e., when there is no excess oxygen in the products and the only C-containing product is \(\mathrm{CO}_{2}\) ). Except in the case of \(\mathrm{N}_{2}\), use only integer coefficients. Including \(\mathrm{N}_{2}\) supplied in the air, write a balanced combustion equation for the complex fuel assuming \(120 \%\) stoichiometric combustion (i.e., when excess oxygen is present in the products and the only C-containing product is \(\mathrm{CO}_{2}\) ). Except in the case of \(\mathrm{N}_{2}\), use only integer coefficients. Calculate the minimum mass (in \(\mathrm{kg}\) ) of air required to completely combust \(1700 \mathrm{~kg}\) of \(\mathrm{C}_{11} \mathrm{H}_{7} \mathrm{~S}\). Calculate the air/fuel mass ratio, assuming \(100 \%\) stoichiometric combustion. Calculate the air/fuel mass ratio, assuming \(120 \%\) stoichiometric combustion.

When ammonia and oxygen are reacted, they produce nitric oxide and water. When \(8.5 \mathrm{~g}\) of ammonia is allowed to react with an excess of \(\mathrm{O}_{2},\) the reaction produces \(12.0 \mathrm{~g}\) of nitrogen monoxide. What is the percentage yield of the reaction?

Iron in the form of fine wire burns in oxygen to form iron(III) oxide. $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ How many moles of \(\mathrm{O}_{2}\) are needed to produce \(7.38 \mathrm{~mol}\) \(\mathrm{Fe}_{2} \mathrm{O}_{3} ?\)

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