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Balance the following equations: (a) \(\mathrm{PCl}_{5}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{POCl}_{3}(l)+\mathrm{HCl}(a q)\) (b) \(\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(I)+\mathrm{NO}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{HI}(s)\) (d) \(\operatorname{Fe}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (e) \(\mathrm{Na}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\) (f) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (g) \(\mathrm{P}_{4}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{3}(l)\) (h) \(\mathrm{PtCl}_{4}(s) \longrightarrow \operatorname{Pt}(s)+\mathrm{Cl}_{2}(g)\)

Short Answer

Expert verified
(a) PCl_5(s) + 2H_2O(l) \rightarrow POCl_3(l) + 4HCl(aq), (b) Cu(s) + 4HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + 2H_2O(l) + 2NO(g), (c) H_2(g) + I_2(s) \rightarrow 2HI(s), (d) 4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s), (e) 2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g), (f) (NH_4)_2Cr_2O_7(s) \rightarrow Cr_2O_3(s) + 2N_2(g) + 4H_2O(g), (g) P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(l), (h) 2PtCl_4(s) \rightarrow 2Pt(s) + 4Cl_2(g).

Step by step solution

01

Balance the Phosphorus and Chlorine in (a)

Start by observing the number of phosphorus and chlorine atoms on both sides of the equation. The phosphorus atoms are balanced, with one on each side. Balance the chlorine atoms by placing a coefficient of 2 in front of \(\mathrm{HCl}\). The revised equation is: \(\mathrm{PCl}_{5}(s) + \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{POCl}_{3}(l) + 2\mathrm{HCl}(a q)\).
02

Balance the Hydrogen in (a)

Now balance the hydrogen atoms by placing a coefficient of 2 in front of \(\mathrm{H}_{2} \mathrm{O}\). The balanced equation is: \(\mathrm{PCl}_{5}(s) + 2\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{POCl}_{3}(l) + 4\mathrm{HCl}(a q)\).
03

Balance Copper in (b)

Copper (\(\mathrm{Cu}\)) is already balanced with one atom on each side of the equation.
04

Balance Nitrogen and Hydrogen in (b)

Place a coefficient of 2 in front of \(\mathrm{HNO}_{3}\) to balance nitrogen, which gives us two nitrogen and two hydrogen atoms on each side. The equation now reads: \(\mathrm{Cu}(s) + 2\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{NO}(g)\).
05

Balance Oxygen in (b)

Notice that by placing a coefficient of 3 in front of \(\mathrm{HNO}_{3}\) and 2 in front of \mathrm{NO}, the oxygen atoms are balanced. The final balanced equation is: \(\mathrm{Cu}(s) + 4\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l) + 2\mathrm{NO}(g)\).
06

Balance (c) Using Simplest Ratio

The elements are already in the simplest whole number ratio 1:1:2. The balanced equation is: \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(s) \longrightarrow 2\mathrm{HI}(s)\).
07

Find the Best Starting Point for (d)

Begin by balancing the iron (\(\mathrm{Fe}\)) by putting a coefficient of 4 in front of it: \(4\mathrm{Fe}(s) + \mathrm{O}_2(g) \longrightarrow \mathrm{Fe}_2\mathrm{O}_3(s)\).
08

Balance the Oxygen in (d)

Balance the oxygen by placing a coefficient of 3 in front of \(\mathrm{O}_2\), giving 6 oxygen atoms on both sides. The balanced equation is: \(4\mathrm{Fe}(s) + 3\mathrm{O}_2(g) \longrightarrow 2\mathrm{Fe}_2\mathrm{O}_3(s)\).
09

Balance Sodium in (e)

Start by balancing the sodium atoms by placing a coefficient of 2 in front of \(\mathrm{NaOH}\) and \(\mathrm{H}_{2}\). The equation now becomes: \(2\mathrm{Na}(s) + 2\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{NaOH}(a q) + \mathrm{H}_{2}(g)\).
10

Balance (f)

Observe that chromium and water are already balanced. Place a coefficient of 4 in front of \(\mathrm{H}_{2}\mathrm{O}\) and a coefficient of 2 in front of \(\mathrm{N}_{2}\) to balance nitrogen and hydrogen. Final balance: \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}(s) \longrightarrow \mathrm{Cr}_{2}\mathrm{O}_{3}(s) + 2\mathrm{N}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(g)\).
11

Balance (g)

Begin by balancing the phosphorus. Place a coefficient of 6 in front of \(\mathrm{PCl}_{3}\) to balance the phosphorus atoms. The equation becomes: \(1\mathrm{P}_{4}(s) + \mathrm{Cl}_{2}(g) \longrightarrow 6\mathrm{PCl}_{3}(l)\). Now balance the chlorines by placing a coefficient of 9 in front of \(\mathrm{Cl}_{2}\). The final equation is: \(\mathrm{P}_{4}(s) + 6\mathrm{Cl}_{2}(g) \longrightarrow 4\mathrm{PCl}_{3}(l)\).
12

Decomposition of (h)

Platinum is already balanced with one atom on each side. Balance the chlorine by placing a coefficient of 2 in front of \(\mathrm{PtCl}_{4}\) and a coefficient of 1 in front of \(\mathrm{Cl}_{2}\). The balanced equation is: \(2\mathrm{PtCl}_{4}(s) \longrightarrow 2\operatorname{Pt}(s) + 4\mathrm{Cl}_{2}(g)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's a mathematical way of describing the conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. In practice, stoichiometry involves using balanced chemical equations to calculate the amounts of reactants needed or products formed.

For instance, if you're baking cookies and the recipe calls for two eggs for every cup of flour, stoichiometry is like the 'recipe' for chemical reactions, telling you how much of each reactant you need to end up with a certain amount of product. In balancing chemical equations, we ensure that the number of atoms for each element in the reactants side is equal to that in the products side. This balancing act reflects the stoichiometric relationship and is essential for any further calculations related to the reaction.
Chemical Reaction
A chemical reaction is a process where reactants transform into products through breaking and forming of chemical bonds. The substances involved can change states, colors, or temperatures, and new substances with different properties from the initial reactants emerge.

Understanding chemical reactions is crucial because they are the fundamental processes that allow us to create new materials and obtain energy. When writing out a chemical reaction, reactants are usually written to the left of an arrow with products on the right side. For a reaction to be balanced, it must adhere to the law of conservation of mass. This particularly means altering coefficients, not subscripts in the chemical formulas, to ensure that the same number of each type of atom appears on both sides of the equation. The balanced chemical equations provide a visual representation of this conservation.
Reaction Stoichiometry
Reaction stoichiometry describes the quantitative relationship between reactants and products in a chemical reaction. It builds on the concept of stoichiometry by specifically focusing on the ratio in which reactants combine and the ratio of products formed.

Understanding reaction stoichiometry is essential for predicting the outcome of reactions and for scaling reactions up or down, such as in industrial processes or laboratory experiments. Using coefficients derived from the balanced chemical equations, we can convert between the amounts of one reactant or product to another, determine the limiting reactant, and calculate the theoretical yield of a reaction. It's like a detailed map that tells you not only how to get from point A to point B but also tells you how much fuel you’ll need and what you’ll see along the way.

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Most popular questions from this chapter

Which of the postulates of Dalton's atomic theory explains why we can calculate a theoretical yield for a chemical reaction?

What is the concentration of \(\mathrm{NaCl}\) in a solution if titration of \(15.00 \mathrm{mL}\) of the solution with \(0.2503 \mathrm{M} \mathrm{AgNO}_{3}\) requires 20.22 mL of the AgNO_ solution to reach the end point? \(\mathrm{AgNO}_{3}(a q)+\mathrm{NaCl}(a q) \rightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q)\)

A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process. (a) The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide. (b) The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water. (c) Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride. (d) The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water. (e) Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides. (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) BrOH (d) CINO \(_{2}\) (e) \(\mathrm{TiCl}_{4}\) (f) NaH

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain?

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