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

Atoms of three different elements are represented by \(\mathrm{O}, \square,\) and \(\Delta .\) Which compound is left over when three molecules of \(\mathrm{O}\Delta\) and three molecules of \(\square \square \Delta\) react to form \(O \square \Delta\) and \(\mathrm{O} \Delta \Delta ?\)

Short Answer

Expert verified
The compound left over after the reaction is \(O\Delta\). There are 3 \(O\Delta\) molecules left over.

Step by step solution

01

Determine the number of atoms before the reaction

There are three molecules of OΔ and three molecules of □□Δ. Count the number of each element in each molecule and sum them together. - OΔ contains 1 O and 1 Δ, so three molecules of OΔ contain 3 O's and 3 Δ's. - □□Δ contains 2 □ and 1 Δ, so three molecules of □□Δ contain 6 □'s and 3 Δ's. Before the reaction, there are 3 O's, 6 □'s, and 6 Δ's in total.
02

Determine the number of atoms after the reaction

The reaction forms O□Δ and OΔΔ. We need to find out how many of each element is present in these two compounds. - O□Δ contains 1 O, 1 □, and 1 Δ - OΔΔ contains 1 O and 2 Δ's Now let's assume there is "x" molecule of O□Δ and "y" molecule of OΔΔ after the reaction. The total number of atoms after the reaction will be: - O's: 1x + 1y = x + y - □'s: 1x = x - Δ's: 1x + 2y = x + 2y
03

Compare the number of atoms and find the leftover compound

Now we need to compare the number of atoms before and after the reaction. We know that the number of atoms is conserved, meaning the number of atoms before the reaction should be equal to the number of atoms after the reaction. 1. For O's: 3 (before) = x + y (after) 2. For □'s: 6 (before) = x (after) 3. For Δ's: 6 (before) = x + 2y (after) Using the second equation (for □'s), we find that x = 6. This means that there are 6 molecules of O□Δ formed in the reaction. Now we have to find the value of y by using the equations for O's and Δ's. If we substitute x = 6 into the first equation (for O's), we get: 3 = 6 + y y = -3 Since the value of y is -3, this means that there are no Δ's left to form any OΔΔ molecule. So, the compound left over is OΔ (3 OΔ molecules to be precise), as OΔΔ couldn't be formed due to the lack of Δ atoms.

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.

Atoms
Atoms are the building blocks of matter. They are the smallest units that define the chemical elements, like oxygen or hydrogen. Every atom contains a nucleus made up of protons and neutrons, surrounded by electrons that orbit the nucleus. Each atom's identity and properties are determined primarily by the number of protons it has. This count is known as the atomic number.
In chemical reactions, atoms rearrange themselves to form new molecules. They are neither created nor destroyed during these reactions, which is an important principle when balancing chemical equations. In the example problem, the elements are represented as different symbols:
  • \(O\) stands for one element (e.g., oxygen),
  • \(\square\) stands for another, and
  • \(\Delta\) stands for the third element.
Recognizing these atoms helps us to determine the quantities present before and after the reaction.
Molecular Composition
Molecular composition refers to the types and numbers of atoms that make up a molecule. Each molecule can be thought of as a small, distinct package of atoms arranged in specific proportions.
In our exercise, molecules like \(O\Delta\) and \(\square\square\Delta\) have different compositions:
  • \(O\Delta\) is composed of one \(O\) atom and one \(\Delta\) atom.
  • \(\square\square\Delta\) consists of two \(\square\) atoms and one \(\Delta\) atom.
When reactions occur, these molecules might change, forming new substances like \(O\square\Delta\) and \(O\Delta\Delta\). Ensuring the correct molecular composition is crucial for understanding the changes that take place during chemical reactions. We need to make sure that the number of each type of atom is accounted for in the products formed.
Conservation of Mass
The conservation of mass is a fundamental principle in chemistry, which states that mass cannot be created or destroyed in a chemical reaction. It implies that the total mass of reactants (substances going into a reaction) must equal the total mass of products (substances produced by the reaction).
Applying this principle to the problem, we started by counting all atoms involved. Before the reaction, we had:
  • Three \(O\) atoms,
  • Six \(\square\) atoms,
  • Six \(\Delta\) atoms.
After the reaction, the same quantities of these atoms must exist. We achieve this by ensuring that the equations for each type of atom are balanced:
  • For \(O\) atoms, the total before and after the reaction should remain three.
  • For \(\square\) atoms, it remains six.
  • For \(\Delta\) atoms, it must also remain six.
This consistency confirms our understanding of the conservation of mass, meaning that the system is balanced correctly. This principle helps identify the correct products formed in chemical reactions and determine what's left over when one type of molecule cannot be entirely formed.

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

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). Combustion of \(35.0 \mathrm{mg}\) of the compound produces \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg}\) \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound?

A compound containing only sulfur and nitrogen is \(69.6 \% \mathrm{S}\) by mass; the molar mass is \(184 \mathrm{g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

The compound \(\mathrm{As}_{2} \mathrm{I}_{4}\) is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is \(3.00 \mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triiodide, what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was \(75.6 \%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Balance the following equations: a. \(\operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{S}_{3}(s)\) b. \(\mathrm{NaHCO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{KClO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. \(\mathrm{Eu}(s)+\mathrm{HF}(g) \rightarrow \mathrm{EuF}_{3}(s)+\mathrm{H}_{2}(g)\)

Consider the following unbalanced equation: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)$$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{kg}\) calcium phosphate with \(1.0 \mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\text { by mass)? }\right.\)

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