Question

Balance the following equations: (a) \(\mathrm{CaS}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) (b) \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)+\mathrm{NaCl}(a q)\) (d) \(\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)\)

Short answer

(a) \(2\mathrm{CaS}(s) + 3\mathrm{H}_2\mathrm{O}(l) \rightarrow 1\mathrm{Ca}(\mathrm{HS})_2(aq) + 1\mathrm{Ca}(\mathrm{OH})_2(aq)\) (b) \(4\mathrm{NH}_3(g) + 5\mathrm{O}_2(g) \rightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_2\mathrm{O}(g)\)

Step-by-step solution

Begin by counting the number of atoms for each element on both sides of the equation. On the left side, there is 1 \(\mathrm{Ca}\), 1 \(\mathrm{S}\), and 2 \(\mathrm{H}\) and 1 \(\mathrm{O}\). On the right side, there are 2 \(\mathrm{Ca}\), 3 \(\mathrm{H}\), 2 \(\mathrm{S}\), and 2 \(\mathrm{O}\). Step 2: Balance Calcium

On the left side of the equation, add a 2 as the coefficient for \(\mathrm{CaS}\) to balance Calcium. Now, there are 2 \(\mathrm{Ca}\), 2 \(\mathrm{S}\), and 2 \(\mathrm{H}\) and 1 \(\mathrm{O}\) on the left side of the equation. Step 3: Balance Hydrogen

Add a 3 as the coefficient for \(\mathrm{H}_2\mathrm{O}\) on the left side to balance Hydrogen. Now, there are 2 \(\mathrm{Ca}\), 2 \(\mathrm{S}\), 6 \(\mathrm{H}\), and 3 \(\mathrm{O}\) on the left side of the equation. Step 4: Balance Final Equation

By comparing all the elements, the balanced equation is: \(2\mathrm{CaS}(s) + 3\mathrm{H}_2\mathrm{O}(l) \rightarrow 1\mathrm{Ca}(\mathrm{HS})_2(aq) + 1\mathrm{Ca}(\mathrm{OH})_2(aq)\) (b) Balancing the reaction: \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Step 1: Count Atoms

On the left side, there is 1 \(\mathrm{N}\), 3 \(\mathrm{H}\), and 2 \(\mathrm{O}\). On the right side, there is 1 \(\mathrm{N}\), 2 \(\mathrm{H}\), and 3 \(\mathrm{O}\). Step 2: Balance Nitrogen

Add a 2 as the coefficient for \(\mathrm{NH}_3\) on the left side and \(\mathrm{NO}\) on the right side to balance the Nitrogen. Now, there are 2 \(\mathrm{N}\), 6 \(\mathrm{H}\), and 2 \(\mathrm{O}\) on the left side of the equation. Step 3: Balance Hydrogen

Put a 3 as the coefficient for \(\mathrm{H}_2\mathrm{O}\) on the right side to balance Hydrogen. Now, there are 2 \(\mathrm{N}\), 6 \(\mathrm{H}\), and 2 \(\mathrm{O}\) on the left side and 2 \(\mathrm{N}\), 6 \(\mathrm{H}\), and 6 \(\mathrm{O}\) on the right side. Step 4: Balance Final Equation

Finally, balance Oxygen by having 5 \(\mathrm{O}_{2}\) on the left side of the equation, which gives us: \(4\mathrm{NH}_3(g) + 5\mathrm{O}_2(g) \rightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_2\mathrm{O}(g)\)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in chemistry. It ensures that the same number of atoms for each element are present on both sides of the equation. This is essential because of the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
To balance an equation, follow these steps:

• Identify the number of atoms for each element in both the reactants and products.
• Use coefficients to adjust the numbers, ensuring that you have the same number of every element on each side.
• Check your work to ensure all elements are balanced. If not, adjust as necessary and recheck.

Remember, coefficients are numbers placed in front of compounds to balance the equation and should be whole numbers.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in a chemical reaction. It is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), and it is used to quantify the proportional relationships in a chemical reaction.
In practice:

• Stoichiometry helps chemists make predictions about the masses, volumes, and moles of substances involved in reactions.
• Understanding molar relationships is essential; one mole of a substance contains Avogadro's number (approximately \(6.022 \times 10^{23}\)) of molecules or atoms.
• It requires conversion between moles, mass, and number of particles using the molar mass concept.

Overall, stoichiometry is like the recipe for a chemical reaction, providing precise measurements of each ingredient needed or produced.

Reaction Stoichiometry

Reaction stoichiometry focuses on the quantitative relationships between reactants and products in a balanced chemical equation.
Here’s why it’s important:

• It allows for the calculation of theoretical yields, which predict how much product is expected from given amounts of reactants.
• Accurate stoichiometric calculations can help identify limiting reactants, which determine the maximum amount of product formed.
• Balancing the chemical equation first is essential for any stoichiometry calculations; unbalanced equations will lead to incorrect results.

By understanding reaction stoichiometry, you can perform efficient calculations and predict the outcome of chemical reactions with accuracy, which is vital in experimental and industrial chemistry.