Question

Balance each chemical equation. (a) \(\mathrm{MnO}_{2}(s)+\mathrm{HCl}(a q) \longrightarrow\) \(\mathrm{Cl}_{2}(g)+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{Co}_{2}(g)+\mathrm{CaSiO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \underset{\mathrm{SiO}_{2}}{\longrightarrow}(s)+\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\) (c) \(\mathrm{Fe}(s)+\mathrm{S}(l) \longrightarrow \mathrm{Fe}_{2} \mathrm{~S}_{3}(s)\) (d) \(\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\)

Short answer

The balanced equations are: (a) \( \text{MnO}_2(s) + 4\text{HCl}(aq) \rightarrow \text{Cl}_2(g) + \text{MnCl}_2(aq) + 2\text{H}_2\text{O}(l) \), (b) \( 3\text{CO}_2(g) + 3\text{CaSiO}_3(s) + 3\text{H}_2\text{O}(l) \rightarrow 3\text{SiO}_2(s) + 3\text{Ca}(\text{HCO}_3)_2(aq)\), (c) \( 4\text{Fe}(s) + 3\text{S}(l) \rightarrow 2\text{Fe}_2\text{S}_3(s)\), (d) \( 3\text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2\text{HNO}_3(aq) + \text{NO}(g)\).

Step-by-step solution

Balancing Equation (a)

Write the correct stoichiometric coefficients that balance the atoms on both sides of the equation. For \( \text{MnO}_2(s) + 4\text{HCl}(aq) \rightarrow \text{Cl}_2(g) + \text{MnCl}_2(aq) + 2\text{H}_2\text{O}(l) \), to balance the chlorine atoms, we place a coefficient of 4 in front of \( \text{HCl} \) on the reactant side. This gives us 4 chlorine atoms on each side. The \( \text{Mn} \) and \( \text{O} \) atoms are already balanced, and placing a 2 in front of \( \text{H}_2\text{O} \) balances the hydrogen atoms.

Balancing Equation (b)

Balance the compounds by adjusting the stoichiometry coefficients. The balanced equation is \( 3\text{CO}_2(g) + 3\text{CaSiO}_3(s) + 3\text{H}_2\text{O}(l) \rightarrow 3\text{SiO}_2(s) + 3\text{Ca}(\text{HCO}_3)_2(aq)\). To balance the calcium, carbon, and silicon atoms, we place coefficients of 3 in front of \( \text{CO}_2, \text{CaSiO}_3, \text{H}_2\text{O} \) and \( \text{SiO}_2 \) as well as in front of \( \text{Ca}(\text{HCO}_3)_2 \) on the product side. This yields a balanced equation with 3 atoms of calcium, silicon, and 9 oxygen atoms on each side (accounting for the ones in \( \text{HCO}_3 \) and \( \text{H}_2\text{O} \)).

Balancing Equation (c)

Adjust the stoichiometric coefficients to balance the atoms. The balanced equation is \( 4\text{Fe}(s) + 3\text{S}(l) \rightarrow 2\text{Fe}_2\text{S}_3(s)\). By placing a 4 in front of \( \text{Fe} \) and a 3 in front of \( \text{S} \) on the reactant side, and a 2 in front of \( \text{Fe}_2\text{S}_3 \) on the product side, the number of iron and sulphur atoms is balanced with 8 iron atoms and 6 sulphur atoms on both sides.

Balancing Equation (d)

Determine and place the appropriate stoichiometric coefficients. The balanced equation is \( 3\text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2\text{HNO}_3(aq) + \text{NO}(g)\). Placing a 3 in front of \( \text{NO}_2 \) on the reactant side and placing a 2 in front of the \( \text{HNO}_3 \) on the product side balances the nitrogen atoms. The hydrogen and oxygen atoms are also balanced, with 2 hydrogen and 7 oxygen atoms on each side of the equation.

Key concepts

Stoichiometric Coefficients

Stoichiometric coefficients are integral to understanding chemical reactions. They indicate the number of moles required for each reactant to fully react and the moles of each product formed. In simple terms, they tell us the 'recipe' needed to perform the chemical process.

For example, consider the equation \begin{center} \( \text{MnO}_2(s) + 4\text{HCl}(aq) \rightarrow \text{Cl}_2(g) + \text{MnCl}_2(aq) + 2\text{H}_2\text{O}(l) \) \end{center}Here, the stoichiometric coefficients are the numbers in front of the chemical formulas—1 for \( \text{MnO}_2 \), 4 for \( \text{HCl} \), and so on. These numbers ensure that the equation adheres to the law of conservation of mass, meaning the same number of atoms for each element must exist on both sides of the equation.

Reactants and Products

In a chemical reaction, reactants are the starting materials that undergo a change to form new substances, which are called products. The transformation from reactants to products involves making and breaking chemical bonds.

Take the reaction between iron (Fe) and sulfur (S) to form iron(III) sulfide (Fe2S3). The reactants here are solid iron and liquid sulfur:\begin{center} \( 4\text{Fe}(s) + 3\text{S}(l) \longrightarrow 2\text{Fe}_2\text{S}_3(s) \) \end{center}After the reaction, the products are a different substance altogether—iron(III) sulfide in this case. Balancing the reaction ensures that the number of atoms of each element in the reactants equals the number in the products.

Chemical Reaction Balancing

Balancing a chemical equation is a crucial step in the study of chemistry. It involves adjusting the stoichiometric coefficients to ensure that the number of atoms of each element is the same on both sides of the equation. A balanced equation complies with the law of conservation of mass and charge.

For instance, balancing the reaction of nitrogen dioxide (NO2) and water (H2O) forming nitric acid (HNO3) and nitric oxide (NO) requires changing stoichiometric coefficients:\begin{center} \( 3\text{NO}_2(g) + \text{H}_2\text{O}(l) \longrightarrow 2\text{HNO}_3(aq) + \text{NO}(g) \) \end{center}Here, placing a coefficient of 3 in front of NO2 and coefficients of 2 in front of HNO3 and 1 in front of H2O and NO ensures both mass and charge are balanced. This represents the essence of chemical reaction balancing.

Conservation of Mass

The conservation of mass is a fundamental principle in chemistry, stating that mass is neither created nor destroyed in a chemical reaction. This law dictates that the mass of the reactants must equal the mass of the products, which is why balancing chemical equations is so vital.

Regardless of how substances are transformed during a reaction, the total mass remains constant. For example, the reaction:\begin{center} \( 3\text{CO}_2(g) + 3\text{CaSiO}_3(s) + 3\text{H}_2\text{O}(l) \longrightarrow 3\text{SiO}_2(s) + 3\text{Ca}(\text{HCO}_3)_2(aq) \) \end{center}shows us that each side of the equation has an equal mass, with 3 moles of carbon, calcium, silicon, and water entering the reaction and the same amount of the respective atoms found in the products. This illustrates the conservation of mass and the importance of balancing chemical equations.