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 O_2. 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

The unbalanced formula for this equation is:

Ca_xO_yH_z -----> 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  x  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

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

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  x  2 moles H₂O

     1             1 mole Mg       = 2 moles of H₂O 

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

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)  +  O_2(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 O₂  = 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?

2C₈H_18 (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  x  16 liters CO₂ 

         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) ---> CaCl₂ (s) ₊ H₂O (g)

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= C_1x6H_2x6O_1x6  = C₆H₁₂O₆

What is wrong with this equation:

In the reaction

Mg +2HCl ---> MgCl₂  + H₂

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

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?