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Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(t) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(t)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) f. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\) g. \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

Short Answer

Expert verified
Balanced chemical equations: a. \(2 \mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(t) \rightarrow 2 \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6\mathrm{HNO}_{3}(a q) \rightarrow 2\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3\mathrm{H}_{2} \mathrm{O}(t)\) c. \(4\mathrm{NH}_{3}(g)+5\mathrm{O}_{2}(g) \rightarrow 4\mathrm{NO}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+4\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+5\mathrm{HCl}(g)\) e. \(\mathrm{CaO}(s)+2\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) f. \(2\operatorname{MoS}_{2}(s)+7\mathrm{O}_{2}(g) \rightarrow 2\mathrm{MoO}_{3}(s)+4\mathrm{SO}_{2}(g)\) g. \(\mathrm{FeCO}_{3}(s)+2\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(t)\)

Step by step solution

01

Write the unbalanced equation

The unbalanced equation is: \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(t) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\)
02

Identify the elements that need to be balanced

These are K, O, and H elements in the compounds.
03

Balance the elements

To balance the equation, we must adjust the coefficients of each compound: \[\boxed{2 \mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(t) \rightarrow 2 \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)}\] #b. Balancing the second equation#
04

Write the unbalanced equation

The unbalanced equation is: \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(t)\)
05

Identify the elements that need to be balanced

These are Fe, O, N, and H elements.
06

Balance the elements

To balance the equation, we must adjust the coefficients of each compound: \[\boxed{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6\mathrm{HNO}_{3}(a q) \rightarrow 2\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3\mathrm{H}_{2} \mathrm{O}(t)}\] #c. Balancing the third equation#
07

Write the unbalanced equation

The unbalanced equation is: \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)
08

Identify the elements that need to be balanced

These are N, O, and H elements.
09

Balance the elements

To balance the equation, we must adjust the coefficients of each compound: \[\boxed{4\mathrm{NH}_{3}(g)+5\mathrm{O}_{2}(g) \rightarrow 4\mathrm{NO}(g)+6\mathrm{H}_{2} \mathrm{O}(g)}\] #d. Balancing the fourth equation#
10

Write the unbalanced equation

The unbalanced equation is: \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\)
11

Identify the elements that need to be balanced

These are P, Cl, H, and O elements.
12

Balance the elements

To balance the equation, we must adjust the coefficients of each compound: \[\boxed{\mathrm{PCl}_{5}(l)+4\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+5\mathrm{HCl}(g)}\] #e. Balancing the fifth equation#
13

Write the unbalanced equation

The unbalanced equation is: \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\)
14

Identify the elements that need to be balanced

These are Ca, O, and C elements.
15

Balance the elements

To balance the equation, we must adjust the coefficients of each compound: \[\boxed{\mathrm{CaO}(s)+2\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)}\] #f. Balancing the sixth equation#
16

Write the unbalanced equation

The unbalanced equation is: \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\)
17

Identify the elements that need to be balanced

These are Mo, S and O elements.
18

Balance the elements

To balance the equation, we must adjust the coefficients of each compound: \[\boxed{2\operatorname{MoS}_{2}(s)+7\mathrm{O}_{2}(g) \rightarrow 2\mathrm{MoO}_{3}(s)+4\mathrm{SO}_{2}(g)}\] #g. Balancing the seventh equation#
19

Write the unbalanced equation

The unbalanced equation is: \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)
20

Identify the elements that need to be balanced

These are Fe, C, H, and O elements.
21

Balance the elements

To balance the equation, we must adjust the coefficients of each compound: \[\boxed{\mathrm{FeCO}_{3}(s)+2\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(t)}\]

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Key Concepts

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

Stoichiometry
Stoichiometry is the part of chemistry that calculates the quantities of reactants and products involved in chemical reactions. It's like a recipe for chemistry:
- If you have a certain amount of one ingredient, stoichiometry tells you how much of the other ingredients you need.
- This is crucial for chemical equations because it helps ensure that reactions occur efficiently and without excess waste.
Balancing chemical equations is the first step in using stoichiometry.
You make sure that the number of atoms for each element is the same on both sides of the reaction arrow, as seen when solving the original exercise problems.
For example, if you are given a chemical equation, you must adjust the coefficients of the compounds, ensuring all elements are equal on both sides.
Take the reaction of ammonia and oxygen as an example:- \[4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2 \text{O}(g)\]Here, stoichiometry was applied by adjusting the coefficients so that we have equal nitrogen, oxygen, and hydrogen atoms in reactants and products. Understanding this aspect helps in designing chemical processes and analyzing reactions in laboratories or industries.
Conservation of Mass
The principle of conservation of mass is a fundamental concept in chemistry.
It states that mass cannot be created or destroyed in a chemical reaction. This means the total mass of the reactants equals the total mass of the products.
When balancing chemical equations, this principle ensures that each element has the same mass before and after the reaction.
Consider the reaction:- \[\text{Fe}_2\text{O}_3(s) + 6 \text{HNO}_3(aq) \rightarrow 2 \text{Fe(NO}_3\text{)}_3(aq) + 3 \text{H}_2\text{O}(t)\]In this example, all elements (iron, nitrogen, oxygen, and hydrogen) are precisely balanced, signifying conservation of mass despite the transformation of reactants into products.
This principle is vital for the chemical industry and environmental science, impacting everything from pharmaceutical production to the assessment of chemical spills.
It's an excellent illustration of nature's equilibrium and a foundational pillar in understanding chemical reactions.
Chemistry Education
Chemistry education teaches us how to decipher the language of molecules and reactions.
It offers the tools to break down complex scientific concepts into understandable knowledge.
In classrooms and educational content, students learn the importance of balanced equations and the basic laws of chemistry, like stoichiometry and conservation of mass.
Equations like:- \[\text{PCl}_5(l) + 4 \text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{PO}_4(aq) + 5 \text{HCl}(g)\]are used to illustrate how chemical reactions occur and why balancing them is necessary.
Through experiments and practical exercises, students experience the real-world application of concepts.
This hands-on approach reinforces theoretical learning and enables students to predict reaction outcomes and solve problems numerically. Chemistry education is about building a solid foundation that supports future learning and sparks curiosity about the natural world.
The step-by-step learning ensures students not only memorize facts but also develop the ability to think like scientists.

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Most popular questions from this chapter

Iron oxide ores, commonly a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\), are given the general formula \(\mathrm{Fe}_{3} \mathrm{O}_{4}\). They yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hydrogen. Balance the following equations for these processes: $$ \begin{aligned} \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \end{aligned} $$

Consider the following reaction: $$ 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating \(0.6498 \mathrm{~g}\) of one of the compounds leaves a residue of \(0.6018 \mathrm{~g}\). Heating \(0.4172 \mathrm{~g}\) of the other compound results in a mass loss of \(0.016 \mathrm{~g}\). Determine the empirical formula of each compound.

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) contains \(2.59 \times 10^{23}\) atoms of hydrogen and is \(17.3 \%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and \(65 \mathrm{~g} / \mathrm{mol}\), what amount (moles) of compound is present, and what is the mass of the sample?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. \(\mathrm{A}\) \(752-\mathrm{g}\) sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

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