Call Now to Set Up Tutoring:

(888) 888-0446

MCAT Physical : Stoichiometry and Analytical Chemistry

Study concepts, example questions & explanations for MCAT Physical

Create An Account

All MCAT Physical Resources

8 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 Next →

Example Question #31 : Stoichiometry And Analytical Chemistry

Compounds can be distinguished from each other by using their molecular weights. The molecular weight of a compound depends on the individual atomic weights of the elements and the amount of each element present in the compound. Consider hexane for example. Hexane has a molecular formula of C_6H_1_4 . This means that it has 6 carbon atoms and 14 hydrogen atoms. To calculate the molecular weight of hexane, we can simply look up the molecular weight of carbon and hydrogen from the periodic table, multiply each molecular weight by the number of atoms (6 for carbon and 14 for hydrogen), and sum the two numbers. The molecular weight of an element is always given in $\frac{grams}{moles}$ . One mole is the defined as the number of atoms in twelve grams of carbon-12.

How many atoms of oxygen are found in 1g of oxygen gas?

Possible Answers:

1.9*10^2^4

1.9* 10^2^2

3.8*10^2^2

3.8*10^2^4

Correct answer:

1.9* 10^2^2

Explanation:

To solve this question, we need to first convert grams to moles, then moles to atoms. The molecular weight (MW) of oxygen gas, O_2 is

$MW\:of\:O_2 = 2\left(16\frac{g}{mol}\right) = 32\frac{g}{mol}$

The amount of moles in 1g of oxygen gas is

$moles\: of \: O_2 = (1g)\left(\frac{1mol}{32g}\right) = 0.0312 \: mol$

There are 6.022* 10^2^3 atoms in $1\:mol$ (this is called the Avogadro’s number); therefore, the number of atoms in of O_2 is

$atoms\ O_2=(0.0312\: mol)\left(\frac{6.022* 10^2^3 \: atoms}{1\: mol}\right)= 1.9*10^2^2\: atoms$

Report an Error

Example Question #32 : Stoichiometry And Analytical Chemistry

Consider two carbohydrates A and B. Molecule A is a 6-carbon carbohydrate and has twice as much molecular weight as molecule B. What can you conclude about molecule B?

Possible Answers:

Molecule B has twice as many oxygen atoms as molecule A

None of these

Molecule A has twice as many oxygen atoms as carbon atoms

Oxygen contributes to about $53\%$ of the molecular weight of both molecules A and B

Correct answer:

Oxygen contributes to about $53\%$ of the molecular weight of both molecules A and B

Explanation:

The empirical formula for carbohydrates is C_nH_2_nO_n , where is the number of carbon atoms. The question states that molecule A has 6 carbons; therefore, n=6 and the molecular formula for molecule A is C_6H_1_2O_6 . The molecular weight (MW) of this compound is calculated using the molecular weight of each atom. The MW of each atom is obtained from the periodic table and is multiplied by the number atoms.

$(6) \left(12\frac{g}{mol}\right) + (12)\left(1.008\frac{g}{mol}\right) + (6) \left(16\frac{g}{mol}\right) = 180\frac{g}{mol}$

So, the molecular weight of molecule A is $180\frac{g}{mol}$ . Since it is also a carbohydrate, molecule B will have the same empirical formula as molecule A. Molecule B has half the molecular weight of molecule A; therefore, molecule B must have half the atoms as molecule A. The molecular formula of molecule B is C_3H_6 O_3 . Note that MW of is $90\frac{g}{mol}$ .

The MW contribution of oxygen to each compound is calculated as follows.

%MW of oxygen for molecule A = $\frac{(6)(16\frac{g}{mol})}{180\frac{g}{mol}} = 0.53$

%MW of oxygen for molecule B = $\frac{(3)(16\frac{g}{mol})}{90\frac{g}{mol}} = 0.53$

Therefore, oxygen contributes to 53% of MW in both compounds.

Report an Error

Example Question #1 : Reaction Calculations And Limiting Reagent

What products will be formed by the reaction between copper (II) fluoride and sodium sulfate?

Possible Answers:

Cu₂Na and F₂SO₄

Cu₂SO₄ and NaF

CuS and NaF

CuSO₄ and NaF

Correct answer:

CuSO₄ and NaF

Explanation:

Recognize that this is a double replacement reaction in which ion pairs will switch. After reaction, copper (II) will now be paired with sulfate, and sodium with flouride according to the (unbalanced) equation below. Remember that sulfate has a charge of -2, and flouride of -1.

$CuF_{2 } + Na_{2}SO_{4}\rightarrow CuSO_{4} + NaF$

Report an Error

Example Question #2 : Reaction Calculations And Limiting Reagent

Which represents the correct balanced equation for the reaction between silver (I) nitrate and magnesium hydroxide?

Possible Answers:

$2Ag_{3}N + 3Mg(OH)_{2} \rightarrow Mg_{}3N_{}2 + 6AgOH$

$2Ag_{2}(NO_{3}) + Mg(OH)_{2} \rightarrow 2Ag_{2}(OH) + Mg(NO_{3})_{2}$

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + 2AgOH$

$2Ag(NO_2)+Mg(OH)_2\rightarrow Mg(NO_2)_2+2AgOH$

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + AgOH$

Correct answer:

$2Ag(NO_{3}) + Mg(OH)_{2} \rightarrow Mg(NO_{3})_{2} + 2AgOH$

Explanation:

Silver (I) nitrate is AgNO₃. Recognizing this allows us to eliminate two answer choices, which incorrectly substitute nitrogen (N) for nitrate (NO₃) or balance the molecular charges incorrectly. Of the two remaining choices, only one is balanced correctly.

Report an Error

Example Question #3 : Reaction Calculations And Limiting Reagent

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

How many molecules of carbon monoxide are produced per liter of hydrogen, when using the natural gas method of production at STP?

Possible Answers:

$5*10^{24}\ \text{molecules}$

$6*10^{21}\ \text{molecules}$

$9*10^{21}\ \text{molecules}$

$3*10^{22}\ \text{molecules}$

Correct answer:

$9*10^{21}\ \text{molecules}$

Explanation:

Since the reaction takes place at STP, we can convert liters of hydrogen to moles using a constant.

$1L\ H_2*\frac{1mol}{22.4L}=0.0446mol\ H_2$

Using this value, we can find the moles of carbon monoxide produced and convert to molecules by using Avogadro's number.

$0.0446mol\ H_2*\frac{1mol\ CO}{3mol\ H_2}*\frac{6.022*10^{23}molec}{1mol}=9*10^{21}\ \text{molecules}\ CO$

Report an Error

Example Question #4 : Reaction Calculations And Limiting Reagent

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

Suppose equal masses of iron and methane were available, along with excess water. What is the difference in the amount of hydrogen molecules that can be produced?

Possible Answers:

Iron will produce about 1.3 times more hydrogen molecules

Iron will produce about eight times more hydrogen molecules

Methane will produce about eight times more hydrogen molecules

Both methods will produce equal amounts of hydrogen molecules

Correct answer:

Methane will produce about eight times more hydrogen molecules

Explanation:

To determine which method will produce more hydrogen molecules, convert equal masses of each reactant into molecules of hydrogen gas. The conversion requires us to convert the initial compound to moles, multiply by the molar ratio from the reaction to find moles of hydrogen, and multiply by Avogadro's number to convert to molecules.

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

$1 g Fe * \frac{1 mol Fe}{55.8 g Fe} * \frac{4 mol H_{2}}{3 mol Fe} * \frac{6.022 * 10^{23} molec H_{2}}{mol H_{2}} = 0.144 * 10^{23} molec$

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

$1 g CH_{4} * \frac{1 mol CH_{4}}{16.0 g CH_{4}} * \frac{3 mol H_{2}}{1 mol CH_{4}} * \frac{6.022 * 10^{23} molec H_{2}}{mol H_{2}} = 1.14 * 10^{23} molec$

Dividing these solutions, we see that the methane method produces roughly eight times the amount of hydrogen molecules.

$\frac{1.14*10^{23}}{0.144*10^{23}}=7.9\approx 8$

Report an Error

Example Question #5 : Reaction Calculations And Limiting Reagent

Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:

$3Fe(s) + 4H_{2}O(g) \rightarrow Fe_{3}O_{2}(s) + 4 H_{2}(g)$

Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:

$CH_{4}(g) + H_{2}O(g) \rightarrow CO(g) + 3 H_{2}(g)$

Suppose $7.7 * 10^{20} atoms$ of iron react with 50g of steam to create iron oxide and hydrogen. What is the limiting reagent and how much of the excess reactant remains?

Possible Answers:

Iron is the limiting reagent and 19g of steam will remain

Steam is the limiting reagent and $5.1 * 10^{15} atoms$ of iron will remain

Iron is a catalyst, and will not be consumed

Iron is the limiting reagent and 49.97g of steam will remain

Correct answer:

Iron is the limiting reagent and 49.97g of steam will remain

Explanation:

In order to identify an excess reactant, we use stoichiometry to convert atoms of iron to mass of steam.

$7.7*10^{20}atoms\ Fe*\frac{1mol\ Fe}{6.022*10^{23}atoms}*\frac{4mol\ H_2O}{3mol\ Fe}*\frac{18.0g\ H_2O}{1mol\ H_2O}=0.031g\ H_2O$

We have 50g of steam available, but only require 0.031g to fully react the given iron. Iron is thus the limiting reagent, since it will be fully consumed first.

50g-0.031g=49.97g

There will be 49.97g of excess steam.

Report an Error

Example Question #1 : Reaction Calculations And Limiting Reagent

How many milliliters of 0.5M NaOH are required to react completely with 10g of BaCO₃?

Possible Answers:

400mL

50mL

100mL

200mL

Correct answer:

200mL

Explanation:

First, write a balanced equation for the reaction.

$BaCO_{3} + 2NaOH \rightarrow Ba(OH)_{2}) + Na_{2}CO_{3}$

Next, convert 10g BaCO₃ to milliliters of NaOH.

$10 g BaCO_{3} * (\frac{1 mol BaCO_{3}}{197.3 g BaCO_{3}}) * (\frac{2 mol NaOH}{1 mol BaCO_{3}}) * (\frac{1000 mL NaOH}{0.5 mol NaOH}) = 200 mL$

Report an Error

Example Question #7 : Reaction Calculations And Limiting Reagent

Which of the following produces a $\small 0.1 M$ solution of potassium chloride?

Possible Answers:

Potassium chloride is not soluble in water

Put $\small 147.51g$ of solute into a container and bring the volume to one liter by adding water while stirring

Put $\small 0.746g$ of solute into a container and bring the volume to $\small 100mL$ by adding water while stirring

Add $\small 7.46g$ of solute to $\small 100mL$ of water

Add $\small 74.6g$ of solute to one liter of water

Correct answer:

Put $\small 0.746g$ of solute into a container and bring the volume to $\small 100mL$ by adding water while stirring

Explanation:

We will be looking for a solution that results in one mole of potassium chloride per ten liters of water.

$0.1M=\frac{0.10mol}{1L}$

We will need to find the molecular weight of potassium chloride.

$KCl\rightarrow MW=39.1+35.5=74.6\frac{g}{mol}$

In order to get the desired concentration, we will need to add one-tenth of this amount to one liter of water.

$7.46g*\frac{1mol}{74.6g}=0.1mol$

Our ratio, then is:

$\frac{7.46g}{1L}$

The only answer to follow this ratio is $\small 0.746g$ of potassium chloride in $\small 100mL$ .

$\frac{7.46g\div10}{1L\div 10}=\frac{0.746g}{0.1L}$

Report an Error

Example Question #8 : Reaction Calculations And Limiting Reagent

$AlCl_{3}+3AgNO_{3}\rightarrow\: Al(NO_{3})_{3}+3AgCl$

If you begin with $\small 66.67g$ of aluminum chloride and unlimited silver nitrate, how many grams of silver chloride can be produced?

Possible Answers:

$116g\ AgCl$

$201g\ AgCl$

$186g\ AgCl$

$215g\ AgCl$

$35.8g\ AgCl$

Correct answer:

$215g\ AgCl$

Explanation:

This is a stoichiometry question requiring us to convert between grams, moles, reactants, and products.

Use the periodic table to find the molar masses of the two compounds in question.

$MW_{AlCl_3}=133.5\frac{g}{mol}$

$MW_{AgCl}=143.37\frac{g}{mol}$

We can use the reaction formula to find the ratio of aluminum chloride to silver chloride. In this case, the ratio is 1:3.

$AlCl_{3}+3AgNO_{3}\rightarrow\: Al(NO_{3})_{3}+3AgCl$

Now we can set up a calculation to convert grams of aluminum chloride to grams of silver chloride, making sure that all units cancel appropriately.

$66.67g\ AlCl_3*\frac{1mol\ AlCl_3}{133.5g\ AlCl_3}*\frac{3mol\ AgCl}{1mol\ AlCl_3}*\frac{143.37g\ AgCl}{1mol\ AgCl}=214.80g\ AgCl$

Report an Error

← Previous 1 2 3 4 5 6 Next →

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Tola
Certified Tutor

Indiana University-Bloomington, Bachelor in Arts, Chemistry. Harvard University, Doctor of Medicine, Homeopathic Medicine.

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Chude
Certified Tutor

Queen Mary's University London, Bachelor of Science, Biology, General.

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Yongmao
Certified Tutor

Nankai University, Bachelor of Science, Chemistry. Washington University in St Louis, Doctor of Philosophy, Organic Chemistry.

All MCAT Physical Resources

8 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in San Francisco-Bay Area, GRE Tutors in Atlanta, GRE Tutors in Phoenix, Physics Tutors in Miami, Chemistry Tutors in Philadelphia, Spanish Tutors in Dallas Fort Worth, Statistics Tutors in Philadelphia, SAT Tutors in Phoenix, GRE Tutors in Houston, Algebra Tutors in Washington DC

Popular Courses & Classes

MCAT Courses & Classes in Chicago, SSAT Courses & Classes in Denver, ISEE Courses & Classes in Washington DC, ACT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in New York City, LSAT Courses & Classes in Philadelphia, GRE Courses & Classes in Seattle, Spanish Courses & Classes in Phoenix, ISEE Courses & Classes in San Diego, LSAT Courses & Classes in Phoenix

Popular Test Prep

SAT Test Prep in Washington DC, MCAT Test Prep in Atlanta, GMAT Test Prep in New York City, GRE Test Prep in Denver, GRE Test Prep in Dallas Fort Worth, ISEE Test Prep in San Diego, ISEE Test Prep in Houston, SSAT Test Prep in Denver, ACT Test Prep in Miami, SSAT Test Prep in Atlanta

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

MCAT Physical : Stoichiometry and Analytical Chemistry

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #31 : Stoichiometry And Analytical Chemistry

Example Question #32 : Stoichiometry And Analytical Chemistry

Example Question #1 : Reaction Calculations And Limiting Reagent

Example Question #2 : Reaction Calculations And Limiting Reagent

Example Question #3 : Reaction Calculations And Limiting Reagent

Example Question #4 : Reaction Calculations And Limiting Reagent

Example Question #5 : Reaction Calculations And Limiting Reagent

Example Question #1 : Reaction Calculations And Limiting Reagent

Example Question #7 : Reaction Calculations And Limiting Reagent

Example Question #8 : Reaction Calculations And Limiting Reagent

All MCAT Physical Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: