Question

Balance each of the following chemical equations. a. \(\mathrm{SiCl}_{4}(l)+\mathrm{Mg}(s) \rightarrow \mathrm{Si}(s)+\mathrm{MgCl}_{2}(s)\) b. \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NOCl}(g)\) c. \(\operatorname{MnO}_{2}(s)+\operatorname{Al}(s) \rightarrow \operatorname{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) d. \(\operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)\) e. \(\mathrm{NH}_{3}(g)+\mathrm{F}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{~F}(s)+\mathrm{NF}_{3}(g)\) f. \(\mathrm{Ag}_{2} \mathrm{~S}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Ag}(s)+\mathrm{H}_{2} \mathrm{~S}(g)\) g. \(\mathrm{O}_{2}(g) \rightarrow \mathrm{O}_{3}(g)\) h. \(\mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)\)

Short answer

The balanced chemical equations are: a. \(\mathrm{SiCl}_{4}(l) + 2\mathrm{Mg}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{MgCl}_{2}(s)\) b. \(\mathrm{NO}(g) + \mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{NOCl}(g)\) c. \(3\mathrm{MnO}_{2}(s) + 2\mathrm{Al}(s) \rightarrow 3\mathrm{Mn}(s) + \mathrm{Al}_{2}\mathrm{O}_{3}(s)\) d. \(16\mathrm{Cr}(s) + 3\mathrm{S}_{8}(s) \rightarrow 8\mathrm{Cr}_{2}\mathrm{S}_{3}(s)\) e. \(6\mathrm{NH}_{3}(g) + 3\mathrm{F}_{2}(g) \rightarrow 3\mathrm{NH}_{4}\mathrm{F}(s) + 3\mathrm{NF}_{3}(g)\) f. \(\mathrm{Ag}_{2}\mathrm{S}(s) + \mathrm{H}_{2}(g) \rightarrow 2\mathrm{Ag}(s) + 2\mathrm{H}_{2}\mathrm{S}(g)\) g. \(3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{O}_{3}(g)\) h. \(32\mathrm{Na}_{2}\mathrm{SO}_{3}(aq) + 3\mathrm{S}_{8}(s) \rightarrow 32\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}(aq)\)

Step-by-step solution

Equation a

Given equation: \(\mathrm{SiCl}_{4}(l)+\mathrm{Mg}(s) \rightarrow \mathrm{Si}(s)+\mathrm{MgCl}_{2}(s)\) We observe that the number of Cl and Mg atoms are not balanced. We can balance them as follows: \[\mathrm{SiCl}_{4}(l) + 2\mathrm{Mg}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{MgCl}_{2}(s)\] Now the equation is balanced.

Equation b

Given equation: \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NOCl}(g)\) We can balance the Cl atoms by placing a 2 in front of the product \(\mathrm{NOCl}\): \[\mathrm{NO}(g) + \mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{NOCl}(g)\] Now the equation is balanced.

Equation c

Given equation: \(\mathrm{MnO}_{2}(s)+\mathrm{Al}(s) \rightarrow \mathrm{Mn}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)\) To balance this equation, we place a 3 in front of \(\mathrm{MnO}_2\) and a 2 in front of \(\mathrm{Al}\): \[3\mathrm{MnO}_{2}(s)+2\mathrm{Al}(s) \rightarrow 3\mathrm{Mn}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)\] Now the equation is balanced.

Equation d

Given equation: \(\mathrm{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2}\mathrm{S}_{3}(s)\) To balance this equation, place a 2 in front of Cr, and a \(3/8 = \frac{3}{8}\) in front of \(∆S_8\): \[2\mathrm{Cr}(s) + \frac{3}{8}\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2}\mathrm{S}_{3}(s)\] However, we must use whole numbers in balancing equations, so we multiply the entire equation by 8 to get rid of the fraction: \[8(2\mathrm{Cr}(s) + \frac{3}{8}\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2}\mathrm{S}_{3}(s))\] \[16\mathrm{Cr}(s) + 3\mathrm{S}_{8}(s) \rightarrow 8\mathrm{Cr}_{2}\mathrm{S}_{3}(s)\] Now the equation is balanced.

Equation e

Given equation: \(\mathrm{NH}_{3}(g)+\mathrm{F}_{2}(g) \rightarrow \mathrm{NH}_{4}\mathrm{F}(s)+\mathrm{NF}_{3}(g)\) To balance this equation, we place a 6 in front of \(\mathrm{NH}_3\), a 3 in front of \(\mathrm{F}_2\), a 3 in front of \(\mathrm{NH}_4\mathrm{F}\), and a 3 in front of \(\mathrm{NF}_3\): \[6\mathrm{NH}_{3}(g) + 3\mathrm{F}_{2}(g) \rightarrow 3\mathrm{NH}_{4}\mathrm{F}(s) + 3\mathrm{NF}_{3}(g)\] Now the equation is balanced.

Equation f

Given equation: \(\mathrm{Ag}_{2}\mathrm{S}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Ag}(s)+\mathrm{H}_{2}\mathrm{S}(g)\) To balance this equation, place a 2 in front of \(\mathrm{Ag}\) and \(\mathrm{H}_2\mathrm{S}\): \[\mathrm{Ag}_{2}\mathrm{S}(s) + \mathrm{H}_{2}(g) \rightarrow 2\mathrm{Ag}(s) + 2\mathrm{H}_{2}\mathrm{S}(g)\] Now the equation is balanced.

Equation g

Given equation: \(\mathrm{O}_{2}(g) \rightarrow \mathrm{O}_{3}(g)\) To balance this equation, place a 3 in front of \(\mathrm{O}_2\) and a 2 in front of \(\mathrm{O}_3\): \[3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{O}_{3}(g)\] Now the equation is balanced.

Equation h

Given equation: \(\mathrm{Na}_{2}\mathrm{SO}_{3}(aq)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}(aq)\) To balance this equation, we place a 4 in front of \(\mathrm{Na}_2\mathrm{SO}_3\) and a 4 in front of \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\). Also place a \(3/8 = \frac{3}{8}\) in front of \(S_8\): \[4\mathrm{Na}_{2}\mathrm{SO}_{3}(aq) + \frac{3}{8}\mathrm{S}_{8}(s) \rightarrow 4\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}(aq)\] Again, we must use whole numbers in balancing equations, so we multiply the entire equation by 8 to get rid of the fraction: \[8(4\mathrm{Na}_{2}\mathrm{SO}_{3}(aq) + \frac{3}{8}\mathrm{S}_{8}(s) \rightarrow 4\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}(aq))\] \[32\mathrm{Na}_{2}\mathrm{SO}_{3}(aq) + 3\mathrm{S}_{8}(s) \rightarrow 32\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}(aq)\] Now the equation is balanced.

Key concepts

Stoichiometry

Stoichiometry is the scientific method used to determine the amounts of substances consumed and produced in chemical reactions. It’s akin to a recipe in cooking, outlining the proportions of each ingredient needed to make the final dish. In chemistry, these 'recipes' are called chemical equations, and stoichiometry is the discipline that ensures these equations are proportionally balanced according to the Law of Conservation of Mass.

For students grappling with stoichiometry, visual aids can be invaluable. Imagine atoms as different colored balls, and reactions as instructions on how to rearrange these balls into new patterns. You start and end with the same number of each, ensuring nothing is lost or gained—consistent with the principle that matter cannot be created or destroyed in a closed system.

The 'mole' concept is critical here; it is the unit of measurement in chemistry that allows chemists to count atoms and molecules by weighing them. Stoichiometry relies on mole ratios derived from the coefficients of the balanced equation. This tells us how many moles of each reactant are needed to produce a certain amount of product, laying the foundation to calculate reactant or product quantities, either at the molecular level or scalable to industrial applications.

Chemical Reactions

Chemical reactions are the processes where substances, referred to as reactants, transform into new substances—or products—as a result of the breaking and forming of chemical bonds. Think of it as a dance, where partners (atoms or molecules) switch and form new pairings according to the rhythm (reaction conditions) of the music (energy changes). Each step of the dance can be represented using symbolic chemical equations which summarize the transformation.

Students may find the concept of chemical reactions easier to understand by categorizing them. Basic reaction types like synthesis, decomposition, single replacement, and double replacement help in predicting the products of an unknown reaction. It's also crucial to know that reactions can be exothermic, releasing energy, or endothermic, requiring an input of energy. Moreover, the speed at which a chemical reaction occurs—the reaction rate—can be influenced by various factors including temperature, concentration, and the presence of catalysts. Being able to write and balance the chemical equations for these reactions is a fundamental skill because it reflects the accuracy in predicting the outcome of chemical processes.

Law of Conservation of Mass

The Law of Conservation of Mass is a fundamental concept in chemistry, stating that mass is neither created nor destroyed in a chemical reaction. This principle is crucial when balancing chemical equations, which is a skill akin to solving a mathematical puzzle while adhering to this law. In essence, what you start with in a reaction must equal what you end up with—albeit in a different form.

When students balance chemical equations, they are ensuring that the number of atoms for each element is the same on both sides of the reaction. This accounting of atoms reflects the indelible truth that matter is persistent. The elegance of the Law of Conservation of Mass is in its universal application; whether a reaction takes place in a test tube or a beaker, an industrial vat, or within the natural environment, the mass of the system will remain constant. Balancing equations is more than just an academic exercise—it embodies a profound concept that ties into the very basics of the physical world.