Balance each chemical equation. (a) \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow \mathrm{NaNO}_{3}(a q)+\mathrm{CuS}(s)\) (b) \(\mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{FeS}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\)

Short Answer

Expert verified
(a) \(\mathrm{Na}_2 \mathrm{S}(aq)+\mathrm{Cu}\left(\mathrm{NO}_3\right)_2(aq) \longrightarrow 2\mathrm{NaNO}_3(aq)+\mathrm{CuS}(s)\) (b) \(4\mathrm{HCl}(aq)+\mathrm{O}_2(g) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)+2\mathrm{Cl}_2(g)\) (c) \(2\mathrm{H}_2(g)+\mathrm{O}_2(g) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)\) (d) \(\mathrm{FeS}(s)+2\mathrm{HCl}(aq) \longrightarrow \mathrm{FeCl}_2(aq)+\mathrm{H}_2 \mathrm{~S}(g)\)

Step by step solution

01

Balance Equation (a)

To balance the equation \(\mathrm{Na}_2 \mathrm{~S}(aq)+\mathrm{Cu}\left(\mathrm{NO}_3\right)_2(aq) \longrightarrow \mathrm{NaNO}_3(aq)+\mathrm{CuS}(s)\), start by counting the number of each type of atom on both sides of the reaction. There are 2 sodium atoms on the left and 1 on the right, so add a coefficient of 2 in front of \(\mathrm{NaNO}_3\) on the right side. Next, verify that the numbers of copper, sulfur, nitrogen, and oxygen atoms are balanced. The copper and sulfur atoms are already balanced. There are 2 nitrates on the left and now 2 on the right, however, this yields a total of 6 oxygen atoms on the left and only 6 on the right which are balanced. The final balanced equation is \(\mathrm{Na}_2 \mathrm{S}(aq)+\mathrm{Cu}\left(\mathrm{NO}_3\right)_2(aq) \longrightarrow 2\mathrm{NaNO}_3(aq)+\mathrm{CuS}(s)\).
02

Balance Equation (b)

In the equation \(\mathrm{HCl}(aq)+\mathrm{O}_2(g) \longrightarrow \mathrm{H}_2\mathrm{O}(l)+\mathrm{Cl}_2(g)\), start by balancing the hydrogen atoms. Add a coefficient of 2 in front of \(HCl\) and \(H_2O\) to balance the hydrogens. This gives us 4 hydrogen atoms on each side. Now balance the chlorine atoms by placing a coefficient of 2 in front of \(\mathrm{Cl}_2\) on the products side. Lastly, to balance the oxygens, balance the \(O_2\) molecules by placing a coefficient of 0.5 in front of \(\mathrm{O}_2\) (since fractional coefficients are typically not used, we can multiply the entire equation by 2 to get integer coefficients). The balanced equation is \(4\mathrm{HCl}(aq)+\mathrm{O}_2(g) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)+2\mathrm{Cl}_2(g)\).
03

Balance Equation (c)

For the combustion reaction of hydrogen \(\mathrm{H}_2(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{H}_2\mathrm{O}(l)\), balance the hydrogen atoms by placing a coefficient of 2 in front of \(\mathrm{H}_2\mathrm{O}\) on the products side. This gives us 4 hydrogen atoms on each side. To balance the oxygen atoms, place a coefficient of 1 in front of \(\mathrm{O}_2(g)\) on the reactant side. This gives us a total of two oxygen atoms on both sides. The balanced equation is now \(2\mathrm{H}_2(g)+\mathrm{O}_2(g) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)\).
04

Balance Equation (d)

In the equation \(\mathrm{FeS}(s)+\mathrm{HCl}(aq) \longrightarrow \mathrm{FeCl}_2(aq)+\mathrm{H}_2 \mathrm{~S}(g)\), we observe that iron (Fe) and sulfur (S) are already balanced. To balance the hydrogen (H) and chlorine (Cl), we place a coefficient of 2 in front of \(\mathrm{HCl}\) on the reactant side, because we have 2 chlorine atoms in \(\mathrm{FeCl}_2\) on the products side. This also balances hydrogen since there are 2 hydrogen atoms on both sides now. The balanced equation becomes \(\mathrm{FeS}(s)+2\mathrm{HCl}(aq) \longrightarrow \mathrm{FeCl}_2(aq)+\mathrm{H}_2 \mathrm{~S}(g)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Stoichiometry
At the heart of chemistry lies stoichiometry, a branch of chemical science focused on quantitatively analyzing the proportions and amounts of substances involved in chemical reactions. It's the arithmetic behind chemistry, allowing scientists and students to calculate the amounts of reactants needed or products formed in a chemical equation.

Stoichiometry is grounded in the balancing of chemical equations, as it uses the coefficients in a balanced equation to perform calculations. For instance, the coefficient numbers indicate the ratio in which reactants combine and products form. When balancing the chemical equation for the reaction of sodium sulfide with copper(II) nitrate, we adjusted coefficients to adhere to these ratios, ensuring the atoms on both sides of the equation correspond and the mass remains consistent throughout the reaction.

In educational terms, think of stoichiometry like a recipe. Just as you need a specified amount of each ingredient to bake a cake, a chemical reaction requires precise amounts of each reactant to proceed as expected and yield the desired product. If we have too much or too little of one component, the reaction won't go to completion, or there will be leftover reactants.
Chemical Reactions
A chemical reaction is a process where substances, known as reactants, transform into different substances called products. This transformation involves breaking and forming chemical bonds, leading to changes in the composition and properties of the involved substances. Reactions can vary greatly in terms of their complexity and the types of changes they involve.

For example, when hydrogen reacts with oxygen, the result is water – a simple yet fundamental reaction critical to life on Earth. As per the provided exercise, balancing the chemical reaction of hydrogen and oxygen (equation c), ensures that we respect the stoichiometry and the ratios in which these gases react. The balanced equation, which obeys the law of conservation of mass, tells us that two molecules of hydrogen gas react with one molecule of oxygen gas to form two molecules of water.

Understanding chemical reactions is crucial for students as it helps them grasp the concept of how different substances interact with each other, the energy changes that occur during these reactions, and the practical outcomes, such as the formation of new materials or the release of energy.
Law of Conservation of Mass
The law of conservation of mass is a fundamental principle in chemistry stating that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to this law, the mass of the reactants in a chemical reaction must equal the mass of the products.

When students balance chemical equations, they are essentially applying the law of conservation of mass. Take for instance, equation d from the exercise, where iron sulfide reacts with hydrochloric acid to produce iron chloride and hydrogen sulfide. Even though the substances change form during the reaction, if we tally the mass of iron, sulfur, chlorine, and hydrogen atoms before and after, they remain constant.

It's vital to recognize that while atoms rearrange during chemical reactions, the same number of each type of atom must be present throughout. This concept assures students that atoms don't vanish or materialize from nowhere; they simply shift from one form to another, anchoring the notion of chemical reactions to a tangible and measurable foundation. Ensuring the balance of equations makes these invisible atomic transactions visible, highlighting the timeless principle that mass is always conserved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Vinegar forms bubbles when it is poured onto the calcium deposits on a faucet, and some of the calcium dissolves. Has a chemical reaction occurred? Explain your answer.

NO is a pollutant emitted by motor vehicles. It is formed by the reaction: (a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)\) Once in the atmosphere, NO (through a series of reactions) adds one oxygen atom to form \(\mathrm{NO}_{2}\). \(\mathrm{NO}_{2}\) then interacts with UV light according to the reaction: (b) \(\mathrm{NO}_{2}(g) \underset{\mathrm{UV} \text { light }}{\mathrm{NO}(g)}+\mathrm{O}(g)\) These freshly formed oxygen atoms then react with \(\mathrm{O}_{2}\) in the air to form ozone \(\left(\mathrm{O}_{3}\right)\), a main component of smog: (c) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\) Classify each of the preceding reactions \((a, b, c)\) as a synthesis, decomposition, single-displacement, or doubledisplacement reaction.

Balance each chemical equation. (a) \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\) (b) \(\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (c) \(\mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{CO}(g)\) (d) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HCl}(g)\)

Predict the type of reaction (if any) that occurs between each pair of substances. Write balanced molecular equations for each. If no reaction occurs, write NO REACTION. (a) aqueous potassium hydroxide and aqueous acetic acid (b) aqueous hydrobromic acid and aqueous potassium carbonate (c) gaseous hydrogen and gaseous oxygen (d) aqueous ammonium chloride and aqueous lead(II) nitrate

What is a chemical reaction? List some examples.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free