Write the conversion factor that you would use to convert from moles of \(\mathrm{Cl}_{2}\) to moles of \(\mathrm{NaCl}\) in the reaction: $$ 2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s) $$

Short Answer

Expert verified
The conversion factor is 2 moles of \(\text{NaCl}\) per mole of \(\text{Cl}_2\).

Step by step solution

01

Analyze the Balanced Chemical Equation

Look at the balanced chemical equation provided to determine the stoichiometric relationship between reactants and products. The coefficient in front of each compound indicates the number of moles that react or are produced.
02

Identify the Moles of Each Substance in Reaction

From the balanced equation, it can be seen that 1 mole of \(\text{Cl}_2\) reacts to produce 2 moles of \(\text{NaCl}\).
03

Write the Conversion Factor

Based on the stoichiometric relationship, you can write the conversion factor from moles of \(\text{Cl}_2\) to moles of \(\text{NaCl}\) as 2 moles of \(\text{NaCl}\) / 1 mole of \(\text{Cl}_2\).

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.

Balanced Chemical Equations
Understanding balanced chemical equations is crucial in the study of chemistry. These equations provide a snapshot of a chemical reaction where the number of atoms for each element in the reactants equals the number of atoms in the products. This balancing adheres to the Law of Conservation of Mass which states that mass cannot be created or destroyed in a chemical reaction.

In our example, the equation
\[2 \mathrm{Na}(s) + \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\] demonstrates a balanced chemical reaction. The coefficient '2' before sodium (Na) and sodium chloride (NaCl) signifies that two moles of sodium react with one mole of chlorine gas to produce two moles of sodium chloride. If these coefficients were not balanced, it would indicate that the reaction does not align with the Law of Conservation of Mass, which is a fundamental principle in chemistry.
Mole Conversion
Mole conversion is a pivotal concept in chemistry as it allows for the quantification of substances involved in a chemical reaction. The mole is the unit used to express amounts of a chemical substance, and it is defined by Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities (atoms, molecules, ions, etc.).

The step by step solution provided translates the stoichiometric coefficients into a mole ratio. This ratio is used as a conversion factor that correlates the moles of one substance to the moles of another. In our example, \(2 \text{ moles of NaCl} / 1 \text{ mole of }\text{Cl}_2\) provides this conversion. This tells us that for every mole of chlorine gas used, two moles of sodium chloride are produced. Understanding mole conversion is essential for accurately determining the quantities needed or produced in a given chemical reaction.
Stoichiometric Relationship
A stoichiometric relationship is the heart of quantitative chemistry. It is the link that connects the moles of various substances in a chemical reaction, as defined by the balanced chemical equation. These relationships help to determine how much of each reactant is needed to produce a certain amount of product.

In our example, the stoichiometric relationship is expressed as a ratio derived from the balanced equation: for every 1 mole of \(\mathrm{Cl}_{2}\), 2 moles of \(\mathrm{NaCl}\) are produced. When performing a reaction in the lab, this ratio allows you to predict the amount of sodium chloride that can be obtained from a given amount of chlorine gas. This understanding is essential for the precise calculation and scalability of chemical reactions, which has real-world applications in industries such as pharmaceuticals, manufacturing, and environmental science.

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

For the reaction shown, calculate how many moles of \(\mathrm{NH}_{3}\) form when each amount of reactant completely reacts. $$ 3 \mathrm{~N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g) $$ (a) \(5.3 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (b) \(2.28 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (c) \(5.8 \times 10^{-2} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (d) \(9.76 \times 10^{7} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\)

Consider the equation for the combustion of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\), the main ingredient in nail polish remover: $$ \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) } \\ {\Delta H_{\text {nxn }}=-1790 \mathrm{~kJ}} \end{array} $$ If a bottle of nail polish remover contains \(155 \mathrm{~g}\) of acetone, how much heat is released by its complete combustion?

Consider the combustion of propane: $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Balance the reaction. (b) Divide all coefficients by the coefficient on propane, so that you have the reaction for the combustion of 1 mol of propane. (c) \(\Delta H_{\mathrm{rxn}}\) for the combustion of one mole of propane is \(-2219 \mathrm{~kJ}\). What mass of propane would you need to burn to generate \(5.0 \mathrm{MJ}\) of heat? (d) If propane costs about \(\$ 0.67 / \mathrm{L}\) and has a density of \(2.01 \mathrm{~g} / \mathrm{cm}^{3}\), how much would it cost to generate \(5.0 \mathrm{MJ}\) of heat by burning propane?

If the theoretical yield of a reaction is \(0.118 \mathrm{~g}\) and the actual yield is \(0.104 \mathrm{~g}\), what is the percent yield?

Consider the generic reaction between reactants \(\mathrm{A}\) and \(\mathrm{B}\) : $$ 3 \mathrm{~A}+4 \mathrm{~B} \longrightarrow 2 \mathrm{C} $$ If a reaction vessel initially contains \(9 \mathrm{~mol} \mathrm{~A}\) and \(8 \mathrm{~mol} \mathrm{~B}\), how many moles of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) will be in the reaction vessel after the reactants have reacted as much as possible? (Assume \(100 \%\) actual yield.)

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