Question

Complete and balance each gas-evolution equation. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{3}(a q) \longrightarrow\) b. \(\mathrm{HCl}(a q)+\mathrm{KHCO}_{3}(a q) \longrightarrow\) c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{NaHSO}_{3}(a q) \longrightarrow\) \(\mathbf{d .}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\)

Short answer

\(a) 2 \mathrm{HNO}_{3}(aq) + \mathrm{Na}_{2} \mathrm{SO}_{3}(aq) \longrightarrow 2\mathrm{NaNO}_{3}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{SO}_{2}(g)\), \(b) \mathrm{HCl}(aq) + \mathrm{KHCO}_{3}(aq) \longrightarrow \mathrm{KCl}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g)\), \(c) \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(aq) + \mathrm{NaHSO}_{3}(aq) \longrightarrow \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g)\), \(d) \left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(aq) + \mathrm{Ca}\left(\mathrm{OH}\right)_{2}(aq) \longrightarrow \mathrm{CaSO}_{4}(s) + 2\mathrm{H}_{2} \mathrm{O}(l) + 2\mathrm{NH}_{3}(g)\).

Step-by-step solution

Identify the Reaction Type

These are gas-evolution reactions where an acid reacts with a base or a bicarbonate, producing a gas, water, and a salt.

Balance Equation a

For the reaction between \(\mathrm{HNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{3}\), write down the products which include a salt, water, and the gas evolved. Then balance the number of atoms for each element on both sides of the equation. The balanced equation is \(2 \mathrm{HNO}_{3}(aq) + \mathrm{Na}_{2} \mathrm{SO}_{3}(aq) \longrightarrow \mathrm{NaNO}_{3}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{SO}_{2}(g)\).

Balance Equation b

For the reaction between \(\mathrm{HCl}\) and \(\mathrm{KHCO}_{3}\), proceed in a similar way to balance the atoms. The balanced equation is \(\mathrm{HCl}(aq) + \mathrm{KHCO}_{3}(aq) \longrightarrow \mathrm{KCl}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g)\).

Balance Equation c

For the reaction between \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(\mathrm{NaHSO}_{3}\), first write down the likely products and then balance. The balanced equation is \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(aq) + \mathrm{NaHSO}_{3}(aq) \longrightarrow \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g)\).

Balance Equation d

For the reaction between \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) and \(\mathrm{Ca}\left(\mathrm{OH}\right)_{2}\), determine the products and balance the atoms. The balanced equation is \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(aq) + \mathrm{Ca}\left(\mathrm{OH}\right)_{2}(aq) \longrightarrow \mathrm{CaSO}_{4}(s) + 2 \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{NH}_{3}(g)\).

Key concepts

Understanding Acid-Base Reactions

Acid-base reactions are a fundamental type of chemical reaction that occur when an acid and a base interact. In these reactions, the acid donates a proton (H⁺) to the base, or the base accepts a proton from the acid. This exchange of protons results in the formation of water (H₂O) and a salt. A classic example is the neutralization reaction where hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form water and sodium chloride (NaCl).

Gas-evolution reactions are a specific category of acid-base reactions where, in addition to water and a salt, a gas is also produced. These reactions often involve carbonates or bicarbonates reacting with acids, leading to the evolution of carbon dioxide gas (CO₂). Understanding these reactions is crucial for identifying the products and writing balanced chemical equations, as shown in the textbook exercise with reactions involving different acids and bicarbonates.

Balancing Chemical Equations

Chemical equations must be balanced to follow the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. To balance an equation, one should ensure that the number of atoms of each element is the same on both sides of the reaction arrow. Begin by writing the unbalanced equation with the correct formulas for all reactants and products. Then, adjust coefficients—one at a time—to balance the atoms, starting typically with an element that appears in only one reactant and one product.

For instance, in the example exercise where hydrochloric acid reacts with potassium bicarbonate, we start by balancing the chlorine and potassium atoms, which are found in singular forms on each side of the equation. Through iterative adjustments, maintaining the correct stoichiometry for all elements, we achieve a balanced chemical equation reflecting the actual proportions in which substances react.

Chemical Stoichiometry

Chemical stoichiometry deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It is based on the balanced chemical equation and allows for the calculation of the amount of reactants required or products formed in a reaction. The coefficients in the balanced equation represent the molar ratio of the substances involved. For example, when the balanced equation shows a 2:1 ratio between reactant A and reactant B, it indicates that two moles of A will react with one mole of B.

Determining the molar ratios is essential in predicting the outcomes of chemical reactions, as seen in the gas-evolution reactions provided in the exercise. Accurate stoichiometry ensures that calculations for the amount of gases evolved, or the quantity of salts formed, are precise, enabling practical applications such as the design of chemical processes or laboratory experiments.