Question

Write a balanced chemical equation for each reaction. a. Solid lead(II) sulfide reacts with aqueous hydrobromic acid (HBr) to form solid lead(II) bromide and dihydrogen monosulfide gas. b. Gaseous carbon monoxide reacts with hydrogen gas to form gaseous methane \(\left(\mathrm{CH}_{4}\right)\) and liquid water. c. Aqueous hydrochloric acid (HCl) reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas. d. Liquid pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) reacts with gaseous oxygen to form gaseous carbon dioxide and liquid water.

Short answer

a) PbS + 2 HBr → PbBr2 + H2S, b) CO + 3 H2 → CH4 + H2O, c) 4 HCl + MnO2 → MnCl2 + 2 H2O + Cl2, d) C5H12 + 8 O2 → 5 CO2 + 6 H2O.

Step-by-step solution

Construct Equation for Reaction a

For the reaction of solid lead(II) sulfide and aqueous hydrobromic acid, start assembling the reactants and products and then balance the equation. Reactants: PbS (solid lead(II) sulfide) and HBr (hydrobromic acid). Products: PbBr2 (solid lead(II) bromide) and H2S (dihydrogen monosulfide gas). Proposed equation: PbS + 2 HBr → PbBr2 + H2S

Construct Equation for Reaction b

Begin with unbalanced equation for the reaction of carbon monoxide with hydrogen gas and then balance the molecules. Reactants: CO (carbon monoxide) and H2 (hydrogen gas). Products: CH4 (methane) and H2O (water). Proposed equation: CO + 3 H2 → CH4 + H2O

Construct Equation for Reaction c

Combine reactants hydrochloric acid and manganese(IV) oxide and equate to products; after start the balance process. Reactants: HCl (hydrochloric acid) and MnO2 (manganese(IV) oxide). Products: MnCl2 (manganese(II) chloride), H2O (water), and Cl2 (chlorine gas). Proposed equation: 4 HCl + MnO2 → MnCl2 + 2 H2O + Cl2

Construct Equation for Reaction d

Start with unbalanced equation for the combustion of pentane in oxygen. Reactants: C5H12 (pentane) and O2 (oxygen). Products: CO2 (carbon dioxide) and H2O (water). Proposed equation: C5H12 + 8 O2 → 5 CO2 + 6 H2O

Key concepts

Chemical Reaction Balancing

Balancing chemical equations can be a tricky task, but with systematic approach, it becomes manageable. The key to success is understanding that a balanced equation obeys the Law of Conservation of Mass, meaning that atoms are neither created nor destroyed in the reaction; they must be equal on both the reactant and product sides.

For instance, in the reaction of lead(II) sulfide with hydrobromic acid, we start by writing down the unbalanced equation. This step involves listing all reactants and products before proceeding to balancing. Once we outline the equation, we adjust the coefficients - the numbers before the chemical formulas - to balance the number of atoms of each element. In our specific example, the balanced equation is achieved when we have 1 atom of lead, 1 atom of sulfur, 2 atoms of bromine, and 2 atoms of hydrogen on both sides of the equation. Beginners often find it easier to start by balancing elements that appear in only one compound on each side, leaving elements that appear in multiple compounds for last.

The more practice you get, the more intuitive this process becomes. Consistently observing the manner in which compounds interact leads to a clearer understanding of the balancing process.

Stoichiometry

Stoichiometry is a section of chemistry that involves using the relationships between reactants and products in a chemical reaction to determine desired quantitative data. It's essentially the math behind chemistry, enabling you to calculate masses, volumes, and moles involved in a reaction.

Consider the balanced chemical equation for the combustion of pentane: \( C_5H_{12} + 8 O_2 \rightarrow 5 CO_2 + 6 H_2O \) . This equation tells us that 1 mole of pentane reacts with 8 moles of oxygen to produce 5 moles of carbon dioxide and 6 moles of water. Understanding the molar ratios of reactants to products is crucial. It's like a recipe - to make the desired amount of product, we need the correct proportions of ingredients.

With stoichiometry, we can predict the product quantities coming out of a reaction or determine the required quantity of a reactant to produce a certain amount of product. Mastering stoichiometry requires a good understanding of mole concept, molar masses and logically applying these in the context of the balanced equation.

Chemical Equation Writing

Writing chemical equations is the foundation of understanding chemical reactions. This skill involves representing the chemical reaction in the form of symbols and formulas. The first step is identifying the reactants and products involved in the chemical reaction. After that, you write down a skeletal equation with these identified substances.

Next, you'll need to incorporate the states of matter for each compound using notations like (s) for solids, (l) for liquids, (g) for gases, and (aq) for aqueous solutions. For instance, when hydrochloric acid reacts with manganese(IV) oxide, the equation starts as \( HCl_{(aq)} + MnO_2_{(s)} \rightarrow MnCl_2_{(aq)} + H_2O_{(l)} + Cl_2_{(g)} \) with states noted.

Once you have the skeletal equation, the focus shifts to balancing, ensuring that the mass and the charge are equal on both sides of the equation. Remember, a well-written chemical equation provides a clear picture of the chemical change, including information on the relative amounts of substances and their physical state.