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The beautiful red mineral rhodochrosite is manganese(II) carbonate. Write an overall, balanced equation for the reaction of the mineral with hydrochloric acid, and name the products.

Short Answer

Expert verified
The balanced equation is \(\text{MnCO}_3 + 2\text{HCl} \rightarrow \text{MnCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\). Products: manganese(II) chloride, carbon dioxide, water.

Step by step solution

01

Identify Reactants

The reactants in the given problem are manganese(II) carbonate, which has the chemical formula \(\text{MnCO}_3\), and hydrochloric acid, \(\text{HCl}\).
02

Predict Products

When \(\text{MnCO}_3\) reacts with \(\text{HCl}\), the products formed are manganese(II) chloride \(\text{MnCl}_2\), carbon dioxide \(\text{CO}_2\), and water \(\text{H}_2\text{O}\).
03

Write the Unbalanced Equation

Based on the reactants and predicted products, the unbalanced chemical equation is: \[\text{MnCO}_3 + \text{HCl} \rightarrow \text{MnCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\]
04

Balance the Chemical Equation

To balance the equation, ensure that the number of atoms of each element on the reactant side equals the number on the product side. The balanced chemical equation is: \[\text{MnCO}_3 + 2\text{HCl} \rightarrow \text{MnCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\]
05

Name the Products

The products of the reaction are manganese(II) chloride (\(\text{MnCl}_2\)), carbon dioxide (\(\text{CO}_2\)), and water (\(\text{H}_2\text{O}\)).

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is crucial in chemistry as it ensures that the conservation of mass is satisfied. When you write a chemical equation, the number of atoms for each element must be the same on both sides of the reaction. To balance an equation, follow these steps:

1. **List Each Element:** Count the number of each type of atom in the reactants and products.
2. **Adjust Coefficients:** Use coefficients to adjust the number of atoms on each side. For example, if you have two chlorines in the product but only one in the reactants, you may need to add a \(2\) in front of the molecule containing chlorine.
3. **Balance Charge:** If needed, balance ionic charges by adding coefficients to molecules with charge.

In our example, manganese(II) carbonate \((\text{MnCO}_3)\) reacts with hydrochloric acid \(\text{HCl}\). Initially, the equation \(\text{MnCO}_3 + \text{HCl} \rightarrow \text{MnCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\) had too few chlorines in the reactants. Therefore, we balance it by placing a \(2\) in front of \(\text{HCl}\): \[\text{MnCO}_3 + 2\text{HCl} \rightarrow \text{MnCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\] This ensures the conservation of mass and follows the law that atoms are neither created nor destroyed.
Inorganic Chemistry
Inorganic chemistry focuses on inorganic compounds, which include minerals, metals, and organometallic compounds. Inorganic compounds often involve ionic bonding between metals and non-metals.

**Compounds in Our Reaction:**
  • Manganese(II) carbonate \(\text{MnCO}_3\) is a common mineral and represents the carbonate group of minerals.
  • Hydrochloric acid \(\text{HCl}\) is a strong acid used frequently in inorganic synthesis and reactions.
  • Products like manganese(II) chloride \(\text{MnCl}_2\), carbon dioxide \(\text{CO}_2\), and water \(\text{H}_2\text{O}\) are typical products of reactions involving acids and carbonates.

These compounds showcase classic inorganic chemistry reactions where metal carbonates react with acids to form salts, gases, and water. Understanding these reactions helps illustrate processes like mineral weathering, acid-neutralization, and industrial synthesis.
Naming Chemical Compounds
Naming chemical compounds systematically is essential for clear communication in chemistry. Each compound has a unique name that identifies its composition and sometimes its structure.

**Naming Rules in Our Reaction:**
  • **Manganese(II) carbonate \(\text{MnCO}_3\):** Here, "Manganese(II)" indicates the oxidation state of the manganese ion, and "carbonate" refers to the polyatomic ion \(\text{CO}_3^{2-}\).
  • **Hydrochloric acid \(\text{HCl}\):** Derived from hydrogen chloride gas dissolved in water. It is a binary acid, identified by the "hydro-" prefix and "-ic" suffix.
  • **Manganese(II) chloride \(\text{MnCl}_2\):** Again, "Manganese(II)" specifies the +2 oxidation state, and "chloride" designates the anion part \(\text{Cl}^-\).
  • **Carbon dioxide \(\text{CO}_2\):** A simple covalent compound using the name of the elements with appropriate prefixes to indicate the number of atoms, though names like carbon dioxide are exceptions often memorized.
  • **Water \(\text{H}_2\text{O}\):** Because of its ubiquity, water retains its common name, though it could be called "dihydrogen monoxide."

Knowing these naming conventions enables clear identification and communication of complex chemical interactions and aids in writing and interpreting chemical equations correctly.

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Most popular questions from this chapter

Write an equation that describes the equilibrium that exists when the weak acid benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right)\) dissolves in water. Identify each of the four species in solution as either Bronsted acids or Bronsted bases. Does the equilibrium favor the products or the reactants? (In acting as an acid, the \(-\mathrm{CO}_{2} \mathrm{H}\) group supplies \(\left.\mathrm{H}^{+} \text {to } \mathrm{form} \mathrm{H}_{3} \mathrm{O}^{+} .\right)\)

Identify and name the water-insoluble product in each reaction and write the net ionic equation: (a) \(\operatorname{CuCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{CuS}+2 \mathrm{HCl}\) (b) \(\mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{KCl}+\mathrm{CaCO}_{3}\) (c) \(\mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{NaI}(\mathrm{aq}) \rightarrow \mathrm{AgI}+\mathrm{NaNO}_{3}\)

The following reaction can be used to prepare iodine in the laboratory. $$\begin{aligned}2 \mathrm{NaI}(\mathrm{s}) &+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{s}) \rightarrow \\\& \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnSO}_{4}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\end{aligned}$$ (a) Determine the oxidation number of each atom in the equation. (b) What is the oxidizing agent, and what has been oxidized? What is the reducing agent, and what has been reduced? (c) Is the reaction observed product-favored or reactant-favored? (d) Name the reactants and products.

Balance the following equations, and name each reactant and product: (a) \(\mathrm{SF}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{HF}(\ell)\) (b) \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{BF}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{HF}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{aq})\)

For each reaction, write an overall, balanced equation and the net ionic equation. (a) the reaction of aqueous lead(II) nitrate and aqueous potassium hydroxide (b) the reaction of aqueous copper(II) nitrate and aqueous sodium carbonate

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