Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Balance this equation and determine which box represents reactants and which box represents products. $$\mathrm{Sb}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SbCl}_{3}(\mathrm{~g})$$

Short Answer

Expert verified
Reactants: Sb and Cl₂; Product: SbCl₃; Balanced equation: 2Sb + 3Cl₂ → 2SbCl₃.

Step by step solution

01

Identify Each Element

Make a list of all elements found in the chemical equation: Sb (Antimony) and Cl (Chlorine).
02

Count Atoms on Each Side

Count the number of atoms for each element on both sides of the equation. On the left, we have 1 Sb and 2 Cl atoms, and on the right, in SbCl₃, we have 1 Sb and 3 Cl atoms.
03

Balance the Chlorine Atoms

To balance the Cl atoms, place a coefficient of 1.5 (or use fractions if needed for intermediate steps) before Cl₂ on the reactants side so that there are 3 Cl atoms. This gives us: \[\mathrm{Sb}(\mathrm{g})+1.5\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SbCl}_{3}(\mathrm{~g})\].
04

Adjust to Whole Numbers

Multiplying the entire equation by 2 gets rid of any fractions: \[2\mathrm{Sb}(\mathrm{g})+3\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2\mathrm{SbCl}_{3}(\mathrm{~g})\]. Now we have whole numbers and there are 2 Sb and 6 Cl atoms on each side.
05

Verify Atom Balance

Double-check that both Sb and Cl are balanced. On the left, there are 2 Sb and 6 Cl, and on the right, the same number of both types of atoms are present, satisfying the conservation of mass.
06

Identify Reactants and Products

The substances on the left side of the chemical equation (Sb and Cl₂ gases) are reactants, and the substance on the right side (SbCl₃ gas) is the product.

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.

Reactants and Products
In the world of chemistry, chemical reactions involve substances transforming into others. These substances onboard each side of the equation are termed reactants and products. The reactants, present on the left side of the equation, are substances that undergo change. For our example, the reactants are Antimony (\( \text{Sb} \)) and Chlorine (\( \text{Cl}_2 \)) gases. Products, found on the equation's right side, are the end findings of the reaction.
In this exercise, Antimony Trichloride (\( \text{SbCl}_3 \)) gas is the resulting product. This output occurs as the reactants recombine their atoms, forming new bonds and thus, new substances that display different properties compared to the starting elements. Understanding which substances are reactants and products is key to grasping how chemical reactions transform matter.
Conservation of Mass
The principle of the conservation of mass tells us that mass cannot be created or destroyed in a chemical reaction. This fundamental notion implies that whatever mass we start with in the reactants will equal the mass found in the products.
When we balance chemical equations, like the one provided, we ensure equal numbers of atoms exist on both sides, confirming the mass balance. In our exercise, we started with 1 Antimony and 2 Chlorine atoms on the reactant side. To get the equally balanced equation, we adjusted it to 2 Antimony and 6 Chlorine atoms on each side, aligning with conservation principles.
Balancing ensures the atom count remains consistent. By using coefficients to adjust the amounts of each substance, we effectively maintain a stable mass flow across the reaction.
Stoichiometry
Stoichiometry, a core concept in chemistry, deals with the quantitative relationships within a chemical reaction. It relies heavily on mole ratios derived from balanced equations to determine the relationships between reactants and products.
In our given example, balancing the equation results in the stoichiometric ratio 2:3:2 for \( \text{Sb} \): \( \text{Cl}_2 \): \( \text{SbCl}_3 \). This ratio tells us how many units of one material will react with a specific number of units of another to form a product.
The calculated whole number coefficients vividly convey this relationship, ensuring that if we start with 2 Antimony and 3 Chlorine molecules, we will end up forming 2 molecules of Antimony Trichloride. This predictive capability is at the heart of stoichiometry, enabling chemists to calculate required and expected amounts efficiently.

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

Identify the acid and base used to form these salts, and write the overall neutralization reaction in both complete and net ionic form. (a) \(\mathrm{NaNO}_{2}\) (b) \(\mathrm{CaSO}_{4}\) (c) NaI (d) \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Silicon and hydrogen form a series of interesting compounds, \(\mathrm{Si}_{x} \mathrm{H}_{y}\). To find the formula of one of them, a \(6.22-\mathrm{g}\) sample of the compound is burned in oxygen. All of the \(\mathrm{Si}\) is converted to \(11.64 \mathrm{~g} \mathrm{SiO}_{2}\) and all of the \(\mathrm{H}\) to \(6.980 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) Determine the empirical formula of the silicon compound.

Nitrogen monoxide is oxidized in air to give brown nitrogen dioxide. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})$$ Starting with \(2.2 \mathrm{~mol} \mathrm{NO},\) calculate how many moles and how many grams of \(\mathrm{O}_{2}\) are required for complete reaction. Calculate what mass of \(\mathrm{NO}_{2}\), in grams, is produced.

The salt calcium sulfate is sparingly soluble in water with a solubility of \(0.209 \mathrm{~g} / 100 \mathrm{~mL}\) water at \(30^{\circ} \mathrm{C}\). If you stirred \(0.550 \mathrm{~g} \mathrm{CaSO}_{4}\) into \(100.0 \mathrm{~mL}\) water at \(30^{\circ} \mathrm{C}\) calculate the molarity of the resulting solution. Calculate the mass of \(\mathrm{CaSO}_{4}\) that would remain undissolved.

An unknown solution contains either lead ions or barium ions, but not both. Which one of these solutions could you use to tell whether the ions present are \(\mathrm{Pb}^{2+}\) or \(\mathrm{Ba}^{2+}\) ? Explain the reasoning behind your choice. $$\mathrm{HCl}(\mathrm{aq}), \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}), \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})$$

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