Stoichiometry Flashcard

What is a stoichiometric coefficient?
A stoichiometric coefficient is the number to the left of a substance in a chemical equation

Fill in the blanks:

 

In the equation: 2 C6H10 + 17 O2 à 12 CO2 + 10 H2O

 

2 moles of C6H10 react with (a)___ moles of O2 —–>

to make (b)____ moles of CO2 and (c)____ moles of H2O.

 

 

 

A. 17

 

B. 12

 

C. 10

 

2 C6H10 + 17 O2 à 12 CO2 + 10 H2O

 

A. In this chemical reaction, what would happen if C6H10 is the limiting reactant and we add more O2. Would your product increase?

 

B. In this chemical reaction, what would happen if C6H10 is the limiting reactant and we add more C6H10. Would it be possible for your product to increase?

A. No, because the C6H10 is the limiting reactant and increasing the O2 would not increase our product.

 

b. Yes, the product could increase now since we increased the limiting reactant.

Study Gay-Lussac’s Law for the test!

;

Know the limitations of this law

Answer is in the book
;A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H.; What is the empirical formula of the compound?

The unbalanced formula for this equation is:

CaxOyHz —–> Ca + O + H

We can now determine how many moles of each product was made:

 

13.5 g Ca  x 1 mole Ca 

     1           40.1 g Ca       =  0.337 moles of Ca

 

 

10.8 g of O  1 mole of O

      1             16 g  O           = 0.675 moles of O

 

 

0.675 g H    x  1 mole of H

      1             1.01 g  H           = 0.668 moles of H 

 

 

Now divide all of them by the smallest to make intergers:

 

0.337

0.337 = 1 mole of Ca

 

0.675

0.337  = 2 moles of O

 

0.668

0.337   = 1.98 = 2 moles of H

 

 

The chemical equation is CaO2H2 which is the empircal formula.

 

So the equation would be:

 

 

CaO2H2 —-> Ca + 2O + 2H

 

 

 

 

 

 

 

 

 

A sample contains 27.1 g of calcium oxide.  How many moles of calcium oxide are in the sample?

Ca has 40.8 amu

O has  16 amu

 

 

27.1 G of CaO  x   1 mole of CaO

      1                 56.08 g of CaO   = 0.483 moles

 Mg(OH)2 is the active ingredient in many laxatives. It can be made with the following reactions
 
Mg (s) +  2H2O (l) —> Mg(OH)2 (s) + H2 (g)
 
If a laxative manufacturer has 25 grams of Mg, how many grams of water must be used to have all of the Mg react to form Mg(OH)2?

This problem asks us to relate the amount of one reactant to the amount of another. This is easy, as long as we start with moles, not grams so we must convert the grams to moles first:

 

25 g Mg x 1 mole of Mg

     1         24.3 g Mg         = 1.0 moles of Mg

 

 

Now we convert the moles of Mg to moles of water:

 

1 mole Mg 2 moles H2O

     1             1 mole Mg       = 2 moles of H2

 

Now we know how much water is needed, we just need to convert from moles to grams:

 

2 moles H2O  x  18 g H2O

        1             1 mole H2O   = 36 grams of H2O

 

So 36 grams of H2O must be used

         

 

 

Is Cu3S3O12 an empirical formula? If not, convert to an empirical formula

Cu3S3O12 is not an empirical formula.

 

To get an empirical formula you must divide by the largest common denominator of the subscipts to get the lowest possible subscripts.

 

For Cu3S3O12 I would divide the subscripts by 3 and then it would an empirical formula:

 

CuSO4

 

 A chemist performs the following reaction to make a metal oxide:

 

2Mg(s)  +  O2(g)  →    2MgO(s)

 

 

If you have 14.8 grams of Mg, what is the minimum number of grams needed of O2?

 

 

 

 

First. we convert Mg to moles:

 

14.8 g Mg x 1 mole of Mg

     1           24.3 g Mg          =  0.609 moles of Mg

 

 

Now convert moles of Mg to moles of O2:

 

0.609 mole of Mg  x 1 mole of O2

         1                 2 moles of Mg =  0.305 moles of O2

 

Now convert moles of O2 to grams:

 

0.305 mole of O2   x  32 g O2 

        1                   1 mole of O= 9.76 grams of O2

 

9.76 grams of oxygen are needed

 

 

 

Example 6.5 in your book:

 

An automobile engine burns gaseous octane (C8H18). If an automobile completely combusts  12 liters of gaseous octance in excess oxygen, how many liters of carbon dioxide will be produced?

 

2C8H18 (g) + 25O2 (g) —> 16CO2 (g) + 18H2O (g)

 

All the substances in the equation are in gas form, we can use Gay-Lussac’s Law to relate their volumes:

 

2 liters of C8H18  =  16 liters of CO2

 

Now we use this relationship to convert:

 

12 liters of C8H18  16 liters CO

         1                  2 liters of  C8H18 = 96 liters of CO2

 

 

So burning 12 liters of octance in excess oxygen will make 96 liters of carbon dioxide.

 

 

Why is the following problem unsolvable?

 

If you have the following reaction, how many liters of H2O would be produced from 2 liters of HCl? 

 

CaO (s) + 2 HCl (g) —> CaCl(s) + H2O (g)

You can’t solve it in liters since the whole equation is not in gas form

From example 6.10 in your book:

 

A compound’s empirical formula is determined by experiment to be CH2O. In a separate experiment, the molecule’s mass was determined to be 180.1 amu. What is the compound’s empirical formula?

 

 

To go from empirical formula to a molecular formula, we must multiply by a common factor.

 

First, figure the mass of CH2O:

1 x 12 amu + 2 x 1.10 amu = 1 x 16 amu = 30 amu

 

Tis isn’t anywhere close to 180.1 amu so you will need to multiply the numbers in the empirical formula by some factor.

 

Here is the formula:

 

common factor = 

 

mass of molecule               =   180.1 amu

mass of empirical formula         30 amu       =  6.00

 

 

Now you multiply each number in the empirical formula by 6 and you will get a molecule whose empirical formula is  CH2O and whose mass is 180.1 amu

 

molecular formula= C1×6H2×6O1×6  = C6H12O6

What is wrong with this equation:

 

In the reaction

 

Mg +2HCl —> MgCl+ H2

 

1 gram of Mg reacts withs 2 grams HCl to make 1 gram of MgCl2  and 1 gram of H2

 

 

The word gram should be replaced with “mole” each time it appears. Stoichiometric coefficients only relate to moles not grams.

Consider the reaction of Mg + 2HCl —> MgCl2 + H2

 

If 4 moles of Mg are reacted with 2 moles of HCL, what is the limiting reactant?

 

 

 According to the equation 2 moles of HCl react with 1 mole of Mg. Thus, when the 2 moles of HCl are exhausted or used up, there will still be leftover Mg. This means HCl is the limiting reactant.

 

 

 

Again consider this formula:

 

Mg + 2HCl —> MgCl2 + H2

 

With the reaction and quantities given in the previous card (4 moles of Mg and 2 moles of HCl) , how many moles of MgCl2 would be formed?

Since HCl is the limiting reactant, we use it in our calculation. According to the equation 2 moles of HCl will make 1 mole of MgCl2

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