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 C_{6}H_{10} + 17 O_{2} à 12 CO_{2} + 10 H_{2}O
2 moles of C_{6}H_{10} react with (a)___ moles of O_{2} —–>
to make (b)____ moles of CO_{2 }and (c)____ moles of H_{2}O.



2 C_{6}H_{10} + 17 O_{2} à 12 CO_{2} + 10 H_{2}O
A. In this chemical reaction, what would happen if C_{6}H_{10} is the limiting reactant and we add more O_{2. }Would your product increase?
B. In this chemical reaction, what would happen if C_{6}H_{10} is the limiting reactant and we add more C_{6}H_{10}. Would it be possible for your product to increase?


A. No, because the C_{6}H_{10} 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 GayLussac’s Law for the test!
;
Know the limitations of this law



;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:
Ca_{x}O_{y}H_{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 CaO_{2}H_{2} which is the empircal formula.
So the equation would be:
CaO_{2}H_{2} —> 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) + 2H_{2}O (l) —> Mg(OH)_{2} (s) + H_{2} (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 x 2 moles H_{2}O
1 1 mole Mg = 2 moles of H_{2}O
Now we know how much water is needed, we just need to convert from moles to grams:
2 moles H_{2}O x 18 g H_{2}O
1 1 mole H_{2}O = 36 grams of H_{2}O
So 36 grams of H_{2}O must be used


Is Cu_{3}S_{3}O_{12 }an empirical formula? If not, convert to an empirical formula 

Cu_{3}S_{3}O_{12 }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 Cu_{3}S_{3}O12 I would divide the subscripts by 3 and then it would an empirical formula:
CuSO_{4}


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 O_{2}?


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 O_{2}:
0.609 mole of Mg x 1 mole of O_{2}
1 2 moles of Mg = 0.305 moles of O_{2}
Now convert moles of O_{2} to grams:
0.305 mole of O_{2}_{ } x 32 g O_{2}
1 1 mole of O_{2 }= 9.76 grams of O_{2}
9.76 grams of oxygen are needed


Example 6.5 in your book:
An automobile engine burns gaseous octane (C_{8}H_{18}). If an automobile completely combusts 12 liters of gaseous octance in excess oxygen, how many liters of carbon dioxide will be produced?
2C_{8}H_{18 (g) }+ 25O_{2} (g) —> 16CO_{2} (g) + 18H_{2}O (g)


All the substances in the equation are in gas form, we can use GayLussac’s Law to relate their volumes:
2 liters of C_{8}H_{18 } = 16 liters of CO_{2}
Now we use this relationship to convert:
12 liters of C_{8}H_{18 }x 16 liters CO_{2 }
1 2 liters of _{ }C_{8}H_{18 }= 96 liters of CO_{2}
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 H_{2}O would be produced from 2 liters of HCl?
CaO (s) + 2 HCl (g) —> CaCl_{2 }(s)_{ + }H_{2}O (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 CH_{2}O. 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 CH_{2}O:
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 CH_{2}O and whose mass is 180.1 amu
molecular formula= C_{1×6}H_{2×6}O_{1×6 } = C_{6}H_{12}O_{6}


What is wrong with this equation:
In the reaction
Mg +2HCl —> MgCl_{2 }+ H_{2}
1 gram of Mg reacts withs 2 grams HCl to make 1 gram of MgCl_{2} and 1 gram of H_{2}


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 —> MgCl_{2} + H_{2}
With the reaction and quantities given in the previous card (4 moles of Mg and 2 moles of HCl) , how many moles of MgCl_{2} 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 MgCl_{2} 
