/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q75E How many molecules of the sweete... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms?

Short Answer

Expert verified

4 molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms

Step by step solution

01

Given data

From the figure it can be identified one Saccharine molecule contains 7 carbon atoms, one nitrogen atom, five hydrogen atoms, three oxygen atoms, one sulfur atom.

02

Determine number of molecules of saccharine

There is total 30 carbon atoms then as each molecule of saccharine contains 7 carbon atoms. So only four molecules of saccharine possible.

There is total 25 hydrogen atoms Also, each molecule of saccharine contains five hydrogen atoms then four molecules contain 20 hydrogen atoms. So, four molecules of saccharine possible.

There is total 12 oxygen atoms Also, each molecule of saccharine contains three oxygen atoms then four molecules contain 12 oxygen atoms. So only four molecules of saccharine possible.

There is a total of 14 nitrogen atom. Each molecule of saccharine contains one nitrogen atoms then four molecules contain 4 nitrogen atoms. So, four molecules of saccharine possible.

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 91Ó°ÊÓ!

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

How many molecules of C2H4Cl2 can be prepared from 15 C2H4molecules and 8 Cl2molecules?

Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H3PO4 react according to the following chemical equation.\(2Cr + 2{H_3}P{O_4} \to 2CrP{O_4} + 3{H_2}\). Determine the limiting reactant.

Indicate what type or types of reaction each of the following represents:

(a) \({H_2}O\left( g \right) + C\left( s \right) \to CO\left( g \right) + {H_2}\left( g \right)\)

(b) \(2KCl{O_3}\left( s \right) \to 2KCl\left( s \right) + 3{O_2}\left( g \right)\)

(c) \(Al{\left( {OH} \right)_3}\left( {aq} \right) + 3HCl\left( {aq} \right) \to AlC{l_3}\left( {aq} \right) + 3{H_2}O\left( l \right)\)

(d) \(Pb\left( {N{O_3}} \right)\left( {aq} \right) + {H_2}S{O_4}\left( {aq} \right) \to PbS{O_4}\left( s \right) + 2HN{O_3}\left( {aq} \right)\)

The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by burning phosphorus in oxygen.

(a) What is the limiting reactant when 0.200 mol of P4 and 0.200 mol of O2 react according to

P4 + 5O2⟶P4O10

(b) Calculate the percent yield if 10.0 g of P4O10is isolated from the reaction.

Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).

(a) Write an equation of the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.

(b)The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solution of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.

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.