Final Chemistry Notebook

STP-Standard Temperature And Pressure, 0 Degree Celsius, 101.3 Atm or 1 atmosphere, 1 mole of gas =22.4 L
Volume to volume: It is the conversion in which we are converting Liter or any other volume to another volume. Liters of an elements into another liter another elements. We do This by starting with liters than to moles; next, mole
of the same element before to moles of another element than mole to the same element and liters and the moles and liter cancel out and all you have to do is divide but for the last you use the 22.4 L because that is the formula because
we are dealing with gases.
Example 1:
For the balanced equation shown below, How much O2 will there be produced if we Combust 3.86 L of Co2?
Co(g) +O2 => 2Co2(g)
1.You first must identify each element’s atomic mass in liters for gas that the problem tells you to do which is 22.4 because it will always be that because that is the formula. You must remember the Formula as shown above STP.
Lastly, we figure out what is equal to what. Identify how many moles each element has and the reaction.
1 mole O2= 2 moles CO2
1 mole Co2= 22.4 Co2 L
mole O2= 22.4 O2 L
2.Next you must set a up a conversion to Convert 3.86 Liters of Co2 to Liters of O2. You must do this by canceling each Other components they have you first cancel Liters then moles than you end with Liters as the result. And use
the highlighted key above to do it.
3.86 L Co2 1 Mol Co2 1 Mol O2 22.4 L O2 = 86.464 L O2
22.4 22.4 L Co2 1 Mol Co2 1 Mol O2 = 44.8 = 1.93 L O2
3.After you set up the Conversion you must multiply everything on the top by itself and everything in the bottom by itself and lastly divide the two by the bottom and you will get your result, but you must put it in Significant figures.
In this example its already in significant figures so we don’t change anything.
2 1.93 L O2
Example 2:
For the balanced equation shown below, assume if we combust 1.3 L of Propane. How much Co2 will be produced?
C3H8 + 5 O2 => 3 Co2+ 4H2O
1.You first must identify each element’s atomic mass in liters for gas that the problem tells you to do which is 22.4 because it will always be that because that is the formula. You must remember the Formula as shown above STP.
Lastly, we figure out what is equal to what. Identify how many moles each element has and the reaction.
1 mole C3H8 = 3 mole Co2
1 mole C3H8 = 22.4 L C3H8
mole Co2 = 22.4 L Co2
2.Next you must set a up a conversion to Convert 1.3 Liters of C3H8 to Liters of Co2. You must do this by canceling each Other components they have you first cancel Liters then moles than you end with Liters as the result. And use
the highlighted key above to do it.
1.3 L C3H8 1 Mol C3H8 3 Mol Co2 22.4 L Co2 = 87.36 L Co2
22.4L C3H8 1 Mol C3H8 1 Mol Co2 = 22.4 = 3.9 L O2
3.After you set up the Conversion you must multiply everything on the top by itself and everything in the bottom by itself and lastly divide the two by the bottom and you will get your result, but you must put it in Significant figures.
In this example its already in significant figures so we don’t change anything.
3.9 L O2
Limiting reagent: is the conversion of grams of an element to mole of the same element than to another element than to grams of that last element Basically grams of and element to another Grans of another elements. Its like mass to
mass but a twist to it you have to do it twice because you are giving 2 different grams and then after you find the answer for both you choose the one with the smaller Number..
Example 1: If 42.2g of C6H5F were reacted with 174g of O2, How many Grams of CO2woul be produced?
1.You first must identify each element’s atomic mass. You find the atomic mass For both grams by looking at the periodic table and need to be rounded you round and multiply by the number of elements it each has the ones you
must add all the products. And figure out what is equal to what. Identify how many moles each element. DO this twice for each gram.
1 mole C6H5F = 96g C6H5F
C6H5F+ 7O2-> 6 Co2
1mole O2= 32g/mol
C 6*12= 72 H5*1= 5 F1*19=19 = 96g/mol O2*16 = 32g/mol C1*12= 12 02*16= 32 = 44g/mol 1mole Co2 = 44g CO2
1 mole C6H5F or 7 moles O2= 6 mole Co2
2. Next you must set a up a conversion to convert 42.2g of C6H5F to CO2 grams of O2 . As well as convert 174g of O2 to grams of CO2. You must do this by canceling each Other
components they have you first cancel grams then moles than you end with grams as the result. And use the highlighted key above to do it.
42.2g C6H5F 1 mol C6H5F 6 mol CO2 44g CO2 = 11140.8
96g C6H5F 1 mol C6H5F 1 mole CO2 = 96 = 116.50
174 g o2 1 mol O2 6 mol CO2 44g CO2 = 45936
32g O2 7 mol O2 1 mole CO2 = 224 = 205.07142
3. After you set up the Conversion you must multiply everything on the top by itself and everything in the bottom by itself and lastly divide the two by the bottom and you will get your result,Choos the one with lowest number and put
in Significant figures.
116 g Co2.
Example 2: IF 5.43 Mg were reacted 27.8 g of H2SO4, how many grams of H2 are produced?
1.You first must identify each element’s atomic mass. You find the atomic mass For both grams by looking at the periodic table and need to be rounded you round and multiply by the number of elements it each has the ones you
must add all the products. And figure out what is equal to what. Identify how many moles each element. DO this twice for each gram.
1 mole H2SO4= 98g H2SO4
Mg+ H2SO4 -> MgSO4+H2
1mole Mg= 24g Mg
Mg 1*24 = 24 g/mol H2*1= 2g/mol H2*1= 2 S1* 32= 32 04*16= 64 = 98g/mol 1mole H2= 2g H2
1 mole H2SO4 or 1 moles Mg= 1 H2
2. Next you must set a up a conversion to convert 5.43g of Mg to grams of H2 . As well as convert 27.8 g of H2SO4 to grams of H2. You must do this by canceling each Other
components they have you first cancel grams then moles than you end with grams as the result. And use the highlighted key above to do it.
5.43g Mg 1 mol Mg 1 mol H2 2g H2 = 10.86
24g Mg 1 mol Mg 1 mole H2 = 24 = .4525
27.8 g H2SO4 1 mol H2SO4 1 mol H2 2g H2 = 55.6
98g H2SO4 1 mol H2SO4 1 mole H2 = 98 = .56734694
3.After you set up the Conversion you must multiply everything on the top by itself and everything in the bottom by itself and lastly divide the two by the bottom and you will get your result, Choose the one with lowest number and
put in Significant figures.
.453 g H2
Percent yield: You must Convert from a element to grams of another element like mass to mass but as well use part over whole, and percent equals actual over theoretician we are solving for theoretical.
Example 1
If the reaction of 27.0 g of O2 produced 59.1% yield how many grams of Co2 would be produced?
C2H3Cl+2O2= =>2 Co2 +3HCl
1.You first must identify each element’s atomic mass. You find the atomic mass by looking at the periodic table and need to be rounded you round and multiply by the number of elements it each has the ones you must add all the
products. And figure out what is equal to what.
1 mole O2= 32 g O2
O2*16=32g C1*12= 12 02*16=32 = 44g
1 mole Co2= 44g/mol Co2
2 mole o2= 2 mole Co2
2. Next you must set a up a conversion to convert 27.0g of O2 to grams of Co2 .. You must do this by canceling each Other
components they have you first cancel grams then moles than you end with grams as the result. And use the highlighted key above to do it.
27.0 g O2 1 mol O2 2 mol Co2 44g CO2 = 1376
32g O2 2 mol O2 1 mole CO2 = 64 = 37.125
3. After you set up the Conversion you must multiply everything on the top by itself and everything in the bottom by itself and lastly divide the two by the bottom and you will get your result, The you use part over whole.
37.125* .591= 21.940875 = 21.9 g CO2
Example 2
If the reaction of 48.3 g of Fe produced 41.4% yield how many grams of Fe3O4 would be produced? 3Fe+4H2O => Fe3O4 +4H2
1.You first must identify each element’s atomic mass. You find the atomic mass by looking at the periodic table and need to be rounded you round and multiply by the number of elements it each has the ones you must add all the
products. And figure out what is equal to what.
1 mole Fe3O4 = 232 g Fe3O4
Fe 1* 56= 56g Fe 3*56= 168 O 4*16=64 = 232g 1 mole Fe= 56 g Fe2
3 mole Fe= 1 mole Fe3O4
2. Next you must set a up a conversion to convert 27.0g of O2 to grams of Co2 .. You must do this by canceling each Other
components they have you first cancel grams then moles than you end with grams as the result. And use the highlighted key above to do it.
48.3g Fe 1 mol Fe 1 mol Fe3O4 232g Fe3O4 = 11205.6
56g Fe 3 mol Fe 1 mole Fe3O4 = 168 = 66.7
3. After you set up the Conversion you must multiply everything on the top by itself and everything in the bottom by itself and lastly divide the two by the bottom and you will get your result, The you use part over whole.
66.7* .414= 27.6138 = 27.6 g Fe3O4
117