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The mineral rhodochrosite [manganese(II) carbonate, \(\left.\mathrm{MnCO}_{3}\right]\) is a commercially important source of manganese. How many moles of \(\mathrm{O}_{2}\) are consumed when one mole \(\mathrm{MnCO}_{3}\) is converted into \(\mathrm{MnO}_{2}\) and \(\mathrm{CO}_{2} ?\)

Short Answer

Expert verified
Answer: One mole of O2 is consumed.

Step by step solution

01

Write a balanced chemical equation

First, we need to write the chemical reaction equation for the given process: the conversion of manganese(II) carbonate (MnCO3) into manganese dioxide (MnO2) and carbon dioxide (CO2). The unbalanced equation can be represented as: MnCO3 + O2 → MnO2 + CO2 Now, we need to balance the equation by adjusting the number of atoms on each side. For this reaction, it is already a balanced equation, and no coefficients need to be adjusted.
02

Use stoichiometry to find the number of moles of O2 consumed

In order to determine the number of moles of O2 that are consumed in the reaction, we need to examine the stoichiometry of the balanced equation. The coefficients of the balanced equation represent the mole ratio of reactants and products: 1 MnCO3 + 1 O2 → 1 MnO2 + 1 CO2 This indicates that one mole of MnCO3 requires one mole of O2 to fully convert it into one mole of MnO2 and one mole of CO2. Therefore, when one mole of MnCO3 is converted, one mole of O2 is consumed.

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is an essential skill in chemistry. It involves ensuring that the number of each type of atom is the same on both sides of a chemical reaction equation. This reflects the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.
To balance a chemical equation, follow these steps:
  • Write down the unbalanced equation. For example, in converting manganese(II) carbonate to manganese dioxide and carbon dioxide, we start with: \( ext{MnCO}_3 + ext{O}_2 ightarrow ext{MnO}_2 + ext{CO}_2 \).
  • Identify each type of atom present and count them on both sides of the equation.
  • Add coefficients to balance each type of atom. The coefficients indicate the number of molecules or moles involved. Remember, you cannot change the subscripts in the chemical formulas as these define the substances.
In our example, the equation is already balanced since there is one manganese (Mn) atom, one carbon (C) atom, and three oxygen (O) atoms on each side. Understanding how to balance equations helps clarify how different substances interact in a reaction.
Mole Ratio
The concept of mole ratio is pivotal in stoichiometry, and it is directly derived from a balanced equation. It allows us to predict the amounts of reactants needed and products formed in a chemical reaction.
In our example, the balanced equation shows that 1 mole of manganese carbonate ( \( ext{MnCO}_3 \) ) reacts with 1 mole of oxygen ( \( ext{O}_2 \) ) to produce 1 mole of manganese dioxide ( \( ext{MnO}_2 \) ) and 1 mole of carbon dioxide ( \( ext{CO}_2 \) ). This gives us a clear mole ratio:
  • 1:1:1:1, which translates to 1 \( ext{MnCO}_3 \) : 1 \( ext{O}_2 \) : 1 \( ext{MnO}_2 \) : 1 \( ext{CO}_2 \).
Using this ratio, if you start with 2 moles of \( ext{MnCO}_3 \), you would need 2 moles of \( ext{O}_2 \) and produce 2 moles each of \( ext{MnO}_2 \) and \( ext{CO}_2 \). The mole ratio makes it straightforward to scale the reaction up or down.
Chemical Reaction Equations
Chemical reaction equations are symbolic representations that describe how substances transform during a chemical reaction. Understanding these equations helps to visualize the entire reaction process.
Every chemical equation needs three major components:
  • Reactants: The starting materials that undergo change. In our exemplar equation, these are manganese carbonate ( \( ext{MnCO}_3 \) ) and oxygen ( \( ext{O}_2 \) ).
  • Products: The substances that are produced as a result of the reaction. Here they are manganese dioxide ( \( ext{MnO}_2 \) ) and carbon dioxide ( \( ext{CO}_2 \) ).
  • The "arrow" notation ( \( ightarrow \) ) separates reactants from products and indicates the direction of the chemical reaction.
Chemical reaction equations are a universal language in chemistry, providing a concise way to convey how molecules interact, rearrange, and transform. They not only depict the substances involved but also offer insights into the energetic and mass relationships within a reaction.

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Most popular questions from this chapter

Some people who prefer natural foods make their own apple cider vinegar. They start with freshly squeezed apple juice that contains about \(6 \%\) natural sugars. These sugars, which all have nearly the same empirical formula, \(\mathrm{CH}_{2} \mathrm{O},\) are fermented with yeast in a chemical reaction that produces equal numbers of moles of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) and carbon dioxide. The product of fermentation, called hard cider, undergoes an acid fermentation step in which ethanol and dissolved oxygen gas react together to form acetic acid (CH \(_{3} \mathrm{COOH}\) ) and water. This acetic acid is the principal solute in vinegar. a. Write a balanced chemical equation for the fermentation of natural sugars to ethanol and carbon dioxide. You may use in the equation the empirical formula given in the preceding paragraph. b. Write a balanced chemical equation for the acid fermentation of ethanol to acetic acid. c. What are the oxidation states of carbon in the reactants and products of the two fermentation reactions? d. If a sample of apple juice contains \(1.00 \times 10^{2} \mathrm{g}\) of natural sugar, what is the maximum quantity of acetic acid that could be produced by the two fermentation reactions?

Silver can be extracted from rocks using cyanide ion. Complete and balance the following reaction for this process: \(\mathrm{Ag}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \rightarrow\) \(\mathrm{Ag}(\mathrm{CN})_{2}-(a q) \quad\) (basic solution)

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Why is \(\mathrm{HSO}_{4}^{-}(a q)\) a weaker acid than \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

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