Question

Balance each of the following chemical equations. a. \(\operatorname{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{CaSO}_{4}(s)+\mathrm{HF}(g)\) b. \(\operatorname{KBr}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HBr}(g)\) c. \(\operatorname{TiCl}_{4}(l)+\operatorname{Na}(s) \rightarrow \operatorname{NaCl}(s)+\operatorname{Ti}(s)\) d. \(\mathrm{K}_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)\) e. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)\) f. \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g) \rightarrow\) \(\mathrm{NaHCO}_{3}(s)+\mathrm{O}_{2}(g)\) g. \(\mathrm{KNO}_{2}(s)+\mathrm{C}(s) \rightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}(g)+\mathrm{N}_{2}(g)\) h. \(\mathrm{BaO}(s)+\mathrm{Al}(s) \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{Ba}(s)\)

Short answer

a. CaF₂(s) + H₂SO₄(l) → CaSO₄(s) + 2HF(g) b. 3KBr(s) + H₃PO₄(aq) → K₃PO₄(aq) + 3HBr(g) c. TiCl₄(l) + 4Na(s) → 4NaCl(s) + Ti(s) d. K₂CO₃(s) → K₂O(s) + CO₂(g) e. KO₂(s) + 2H₂O(l) → 2KOH(aq) + O₂(g) f. Na₂O₂(s) + 2H₂O(g) + 2CO₂(g) → 2NaHCO₃(s) + O₂(g) g. 2KNO₂(s) + C(s) → K₂CO₃(s) + CO(g) + 2N₂(g) h. 3BaO(s) + 2Al(s) → Al₂O₃(s) + 3Ba(s)

Step-by-step solution

a. Balance CaF₂(s) + H₂SO₄(l) → CaSO₄(s) + HF(g)

First, balance the Calcium atoms by placing a coefficient of 1 in front of CaF₂ and CaSO₄, as there is one Calcium atom on each side. CaF₂(s) + H₂SO₄(l) → CaSO₄(s) + HF(g) Next, balance the Fluorine atoms by placing a coefficient of 2 in front of HF, since there are two fluorine atoms in CaF₂. CaF₂(s) + H₂SO₄(l) → CaSO₄(s) + 2HF(g) Finally, check the Hydrogen and Sulfur atoms to ensure that they are balanced. We have two Hydrogen atoms on both sides and one Sulfur atom on both sides, so no additional coefficients are needed: CaF₂(s) + H₂SO₄(l) → CaSO₄(s) + 2HF(g)

b. Balance KBr(s) + H₃PO₄(aq) → K₃PO₄(aq) + HBr(g)

First, balance the Potassium atoms by placing a coefficient of 3 in front of KBr. 3KBr(s) + H₃PO₄(aq) → K₃PO₄(aq) + HBr(g) Next, balance the Bromine atoms by placing a coefficient of 3 in front of HBr. 3KBr(s) + H₃PO₄(aq) → K₃PO₄(aq) + 3HBr(g) Finally, check the Hydrogen and Phosphorus atoms to ensure that they are balanced. We have three Hydrogen atoms on the reactant side and three on the product side, and one Phosphorus atom on both sides, so no additional coefficients are needed: 3KBr(s) + H₃PO₄(aq) → K₃PO₄(aq) + 3HBr(g)

c. Balance TiCl₄(l) + Na(s) → NaCl(s) + Ti(s)

First, balance the Titanium atoms by placing a coefficient of 1 in front of TiCl₄ and Ti, as there is one Titanium atom on each side. TiCl₄(l) + Na(s) → NaCl(s) + Ti(s) Next, balance the Chlorine atoms by placing a coefficient of 4 in front of NaCl, since there are four Chlorine atoms in TiCl₄. TiCl₄(l) + Na(s) → 4NaCl(s) + Ti(s) Finally, balance the Sodium atoms by placing a coefficient of 4 in front of Na. TiCl₄(l) + 4Na(s) → 4NaCl(s) + Ti(s)

d. Balance K₂CO₃(s) → K₂O(s) + CO₂(g)

First, balance the Potassium atoms by placing a coefficient of 1 in front of K₂CO₃ and K₂O. K₂CO₃(s) → K₂O(s) + CO₂(g) Next, balance the Carbon atoms by placing a coefficient of 1 in front of CO₂. K₂CO₃(s) → K₂O(s) + CO₂(g) Finally, check the Oxygen atoms to ensure that they are balanced. We have three Oxygen atoms on both sides, so no additional coefficients are needed: K₂CO₃(s) → K₂O(s) + CO₂(g)

e. Balance KO₂(s) + H₂O(l) → KOH(aq) + O₂(g)

First, balance the Potassium atoms by placing a coefficient of 1 in front of KO₂ and KOH: KO₂(s) + H₂O(l) → KOH(aq) + O₂(g) Next, balance the Oxygen atoms by placing a coefficient of 2 in front of KOH: KO₂(s) + H₂O(l) → 2KOH(aq) + O₂(g) Finally, balance the Hydrogen atoms by placing a coefficient of 2 in front of H₂O: KO₂(s) + 2H₂O(l) → 2KOH(aq) + O₂(g)

f. Balance Na₂O₂(s) + H₂O(g) + CO₂(g) → NaHCO₃(s) + O₂(g)

First, balance the Sodium atoms by placing a coefficient of 2 in front of NaHCO₃: Na₂O₂(s) + H₂O(g) + CO₂(g) → 2NaHCO₃(s) + O₂(g) Next, balance the Carbon atoms by placing a coefficient of 2 in front of CO₂: Na₂O₂(s) + H₂O(g) + 2CO₂(g) → 2NaHCO₃(s) + O₂(g) Finally, balance the Hydrogen and Oxygen atoms by placing a coefficient of 2 in front of H₂O: Na₂O₂(s) + 2H₂O(g) + 2CO₂(g) → 2NaHCO₃(s) + O₂(g)

g. Balance KNO₂(s) + C(s) → K₂CO₃(s) + CO(g) + N₂(g)

First, balance the Potassium atoms by placing a coefficient of 2 in front of KNO₂: 2KNO₂(s) + C(s) → K₂CO₃(s) + CO(g) + N₂(g) Next, balance the Carbon atoms by placing a coefficient of 1 in front of C and CO: 2KNO₂(s) + C(s) → K₂CO₃(s) + CO(g) + N₂(g) Next, balance the Nitrogen atoms by placing a coefficient of 2 in front of N₂: 2KNO₂(s) + C(s) → K₂CO₃(s) + CO(g) + 2N₂(g) Finally, check the Oxygen atoms to ensure that they are balanced. We have six Oxygen atoms on both sides, so no additional coefficients are needed: 2KNO₂(s) + C(s) → K₂CO₃(s) + CO(g) + 2N₂(g)

h. Balance BaO(s) + Al(s) → Al₂O₃(s) + Ba(s)

First, balance the Barium atoms by placing a coefficient of 1 in front of Ba: BaO(s) + Al(s) → Al₂O₃(s) + Ba(s) Next, balance the Aluminum atoms by placing a coefficient of 2 in front of Al: BaO(s) + 2Al(s) → Al₂O₃(s) + Ba(s) Finally, balance the Oxygen atoms by placing a coefficient of 3 in front of BaO: 3BaO(s) + 2Al(s) → Al₂O₃(s) + 3Ba(s)

Key concepts

Stoichiometry

Stoichiometry is the mathematical study of the quantitative aspects of chemical reactions. It is rooted in the principle of the conservation of mass, where the total mass of the reactants equals the total mass of the products in a chemical reaction. To achieve a balanced equation, the number of atoms for each element must be the same on both sides of the reaction. This process requires the use of coefficients, which are whole numbers placed in front of chemical formulas to ensure the number of atoms for each element is conserved.

For example, in the step-by-step solution we see that to balance the reaction involving calcium fluoride (CaF₂) and sulfuric acid (H₂SO₄), we start by balancing one element at a time. This systematic approach, often referred to as the 'hit and trial' or 'tare and trial' method, allows us to adjust coefficients until the number of atoms of each element is equal on both sides of the reaction. In this scenario, we placed a coefficient of 2 in front of HF to balance the fluorine atoms.

Understanding stoichiometry is not just about balancing equations; it also enables calculation of reactant and product quantities, often in moles, in a given reaction. This forms the basis for further calculations, such as determining the limiting reactant, calculating percent yield, or deducing empirical and molecular formulas.

Chemical Reactions

Chemical reactions are processes where reactants convert into products through the breaking and forming of chemical bonds. During a reaction, substances undergo chemical changes, leading to the formation of new substances with different properties. There are several types of chemical reactions, including synthesis, decomposition, single displacement, double displacement, combustion, and acid-base reactions.

The step-by-step solutions provided demonstrate how different types of reactions are balanced. For instance, the decomposition of potassium carbonate (K₂CO₃) into potassium oxide (K₂O) and carbon dioxide (CO₂) is an example of a decomposition reaction. By contrast, the reaction between calcium fluoride and sulfuric acid to form calcium sulfate and hydrogen fluoride exemplifies a double displacement reaction. Each reaction has its unique set of reactants and products which must be balanced according to the laws of conservation of mass and charge.

In order to write and balance chemical equations effectively, understanding the reactants' and products' chemical formulas, as well as the stoichiometry involved, is crucial. Given time and practice, the ability to predict products and balance equations becomes an intuitive skill for students.

Mole Concept

The mole concept is a central theme in chemistry that simplifies the measurement of the amount of substances involved in chemical reactions. It defines a mole as exactly 6.022×10²³ particles (Avogadro's number) of any given substance, which corresponds to the number of atoms found in 12 grams of carbon-12. The use of moles allows chemists to count particles by weighing them, bridging the gap between the macroscopic scale we can measure and the atomic scale where reactions occur.

When balancing equations, as seen in the provided solutions, you are not only ensuring the equality of atoms but also the moles of atoms in both reactants and products. For example, balancing the oxygen atoms in the reaction involving sodium peroxide (Na₂O₂), water (H₂O), and carbon dioxide (CO₂) to form sodium bicarbonate (NaHCO₃) and oxygen gas (O₂) involves an understanding of both the stoichiometry and the mole concept.

The mole concept also facilitates the calculation of reactant or product masses using the molar mass, which is the mass of one mole of a substance (in grams per mole). Mastery of the mole concept is essential for quantitative chemistry and enables students to grasp more complex topics in the subject.