Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 137

The following reaction is stoichiometric as written$$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}+\mathrm{NaOC}_{2}\mathrm{H}_{5} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}+\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}+\mathrm{NaCl}$$ but it is often carried out with an excess of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with \(6.83 \mathrm{~g}\) of \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\), how many grams of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) would be needed to have a 50 percent molar excess of that reactant?

Short Answer

Expert verified
A 50 percent molar excess of \(NaOC_2H_5\) equates to an approximate mass of 7.79 g

Step by step solution

01

Calculate Moles of Given Reactant

The molar mass of \(C_4H_9Cl\) is calculated by adding up the molar masses of its constituent atoms. This results in a molar mass of approximately 108.6 g/mol. Therefore, the amount in moles of \(C_4H_9Cl\) that we have can be calculated using its mass and its molar mass: Moles of \(C_4H_9Cl = \frac{Mass}{Molar Mass} = \frac{6.83 g}{108.6 g/mol} = 0.063 mol
02

Calculate Moles of Required Reactant

From the balanced chemical equation, we can see that the molar ratio of \(NaOC_2H_5\) to \(C_4H_9Cl\) is 1:1. However, we are asked to calculate the amount of \(NaOC_2H_5\) required for a 50% molar excess. Therefore, the moles of \(NaOC_2H_5\) required would be 1.5 times the moles of \(C_4H_9Cl\). Hence, Moles of \(NaOC_2H_5\) required = 1.5 * Moles of \(C_4H_9Cl\) = 1.5 * 0.063 mol = 0.095 mol
03

Calculate Mass of Required Reactant

Once we have the moles of \(NaOC_2H_5\) required, we can then convert it into mass using its molar mass. The molar mass of \(NaOC_2H_5\) is approximately 82 g/mol. Therefore, the mass of \(NaOC_2H_5\) required will be Moles * Molar Mass = 0.095 mol * 82 g/mol = 7.79 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.

Start your free trial

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

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

When heated, lithium reacts with nitrogen to form lithium nitride: $$6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s)$$What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) are heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2} ?\) If the actual yield of \(\mathrm{Li}_{3} \mathrm{~N}\) is \(5.89 \mathrm{~g},\) what is the percent yield of the reaction?

Ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) an important industrial organic chemical, can be prepared by heating hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) at \(800^{\circ} \mathrm{C}:$$$\mathrm{C}_{6} \mathrm{H}_{14} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\text { other products }$$ If the yield of ethylene production is 42.5 percent, what mass of hexane must be reacted to produce \)481 \mathrm{~g}$ of ethylene?

A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When \(X\) is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent \(\mathrm{Mn}\) and 28.0 percent \(\mathrm{O}\) is formed. (a) Determine the empirical formulas of X and Y. (b) Write a balanced equation for the conversion of \(\mathrm{X}\) to \(\mathrm{Y}\)

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) of mass \(13.43 \mathrm{~g}\) is completely burned in oxygen. If the total mass of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(64.84 \mathrm{~g},\) calculate the fraction of \(\mathrm{CH}_{4}\) in the mixture.

Use the formation of water from hydrogen and oxygen to explain the following terms: chemical reaction, reactant, product.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation
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