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Consider the combustion of carbon monoxide (CO) in oxygen gas: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ Starting with 3.60 moles of \(\mathrm{CO},\) calculate the number of moles of \(\mathrm{CO}_{2}\) produced if there is enough oxygen gas to react with all the CO.

Short Answer

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3.60 moles of CO₂ are produced.

Step by step solution

01

Write Down the Balanced Equation

The balanced chemical equation for the combustion of carbon monoxide (CO) in oxygen gas is given as:\[ 2 \mathrm{CO}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) \]
02

Identify the Mole Ratios

From the balanced chemical equation, we see that 2 moles of CO produce 2 moles of CO\(_2\). This gives a mole ratio of \(\frac{2 \text{ moles of } \mathrm{CO}}{2 \text{ moles of } \mathrm{CO}_2}\). Simplified, the mole ratio is 1:1.
03

Calculate Moles of CO2 Produced

Starting with 3.60 moles of CO, and using the 1:1 mole ratio, the moles of CO\(_2\) produced can be calculated as follows:\[ \text{Moles of } \mathrm{CO}_2 = 3.60 \text{ moles of } \mathrm{CO} \times \frac{2 \text{ moles of } \mathrm{CO}_2}{2 \text{ moles of } \mathrm{CO}} = 3.60 \text{ moles of } \mathrm{CO}_2 \]

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Key Concepts

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

Stoichiometry
Stoichiometry plays a crucial role in understanding chemical reactions. It deals with the quantitative relationships between reactants and products in a chemical reaction. In this case, by mastering stoichiometry, you can predict how much product will form from a given amount of reactants.
For example, in the combustion of carbon monoxide to form carbon dioxide, stoichiometry helps us calculate the exact amount of carbon dioxide produced when starting with a certain amount of carbon monoxide.
The balanced chemical equation is at the heart of stoichiometry, acting like a "recipe" for the reaction. In the equation provided: \[2\mathrm{CO}(g) + \mathrm{O}_2(g) \rightarrow 2\mathrm{CO}_2(g)\], each molecule or mole of a compound is symbolized by coefficients that illustrate how many units of each substance are involved.
These coefficients help create mole ratios, which are essential for stoichiometric calculations, allowing you to convert moles of one substance to moles of another using simple ratio calculations. For instance, every 2 moles of CO reacts to produce 2 moles of CO₂, giving you a mole ratio that simplifies your calculations.
Combustion Reactions
Combustion reactions are a type of chemical reaction where fuel reacts with an oxidant, often oxygen, releasing energy, usually in the form of heat and light. These reactions are exothermic, which means they release energy.
The combustion of carbon monoxide (CO) is a classic example of such a reaction, where CO combines with oxygen gas (O₂) to form carbon dioxide (CO₂). This particular reaction can be represented by the balanced chemical equation: \[2\mathrm{CO}(g) + \mathrm{O}_2(g) \rightarrow 2\mathrm{CO}_2(g)\].
Understanding combustion reactions is essential for both practical applications, such as energy production and environmental protection, by ensuring the complete burning of fuels to prevent toxic emissions like CO.
Moreover, these reactions showcase the importance of oxygen in the chemical process. Without sufficient oxygen, incomplete combustion occurs, leading to the formation of harmful substances.
Mole Concept
The mole concept is a central unit in chemistry that helps in quantifying substances. It provides a bridge between the atomic world and the macro-scale quantities we can measure in a laboratory.
A mole is defined as the amount of substance that contains as many elementary entities as there are atoms in 12 grams of pure carbon-12 (\(^{12}C\)). This number is Avogadro's number, approximately \(6.022 \times 10^{23}\).
In the exercise, understanding the mole concept allows you to determine the number of moles of CO₂ produced from a known number of moles of CO. Since 3.60 moles of CO are provided, and the reaction has a 1:1 mole ratio, a straightforward calculation using the mole concept gives the same number of moles of CO₂, which is 3.60 moles.
Essentially, the mole concept enables chemists to work with the submicroscopic amount of substances, making calculations and predictions about chemical reactions more feasible.

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Most popular questions from this chapter

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a component of gasoline. Complete combustion of octane yields \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\). Incomplete combustion produces \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO},\) which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gallon (gal) of octane is burned in an engine. The total mass of \(\mathrm{CO}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(11.53 \mathrm{~kg} .\) Calculate the efficiency of the process; that is, calculate the fraction of octane converted to \(\mathrm{CO}_{2}\). The density of octane is \(2.650 \mathrm{~kg} / \mathrm{gal}\).

When \(0.273 \mathrm{~g}\) of \(\mathrm{Mg}\) is heated strongly in a nitrogen \(\left(\mathrm{N}_{2}\right)\) atmosphere, a chemical reaction occurs. The product of the reaction weighs \(0.378 \mathrm{~g}\). Calculate the empirical formula of the compound containing \(\mathrm{Mg}\) and \(\mathrm{N}\). Name the compound.

A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\). Assuming complete decomposition, calculate the number of grams of \(\mathrm{O}_{2}\) gas that can be obtained from \(46.0 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\). (The products are \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\).)

The compound 2,3 -dimercaptopropanol \(\left(\mathrm{HSCH}_{2} \mathrm{CHSHCH}_{2} \mathrm{OH}\right),\) commonly known as British Anti-Lewisite (BAL), was developed during World War I as an antidote to arsenic-containing poison gas. (a) If each BAL molecule binds one arsenic (As) atom, how many As atoms can be removed by \(1.0 \mathrm{~g}\) of BAL? (b) BAL can also be used to remove poisonous heavy metals like mercury \((\mathrm{Hg})\) and lead \((\mathrm{Pb})\). If each \(\mathrm{BAL}\) binds one \(\mathrm{Hg}\) atom, calculate the mass percent of \(\mathrm{Hg}\) in a BAL-Hg complex. (An \(\mathrm{H}\) atom is removed when a BAL molecule binds an \(\mathrm{Hg}\) atom.)

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