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First find molecular masses of substances. 2. Convert 122.45 g of KClO3 to moles. 3 Find molar ratios from balanced equation.
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Chapter 17
Atomic weight on periodic table is average of natural abundance of isotopes Atomic mass is the number of nucleons in a particular atom—specified by isotope Molecular mass is the mass of one mole of molecules One atomic mass number of grams 6.0221367x10 23 molecules
O atomic weight 15.9996 (round to 16 for this class) O-16 atomic mass 16 u Molecular oxygen O atomic mass 32 uMolecular oxygen O 2 atomic mass 32 u Molecular O 2 molecular mass 32 g/mole
C=12 g/mole, O 2 =32 g/mole CO 2 =12+32=44 g/mole
Na = 23 g/mole C = 12 g/mole O = 16 g/mole 2 N 2 Na, 3 O, so multiply 3 O lti l 2(23) + 12 + 3(16) = 106 g/mole
Cl = 35. Use for some questions
2 KClO 3 2 KCl + 3 O 2
How many grams of oxygen can be produced from 122.45 g of KClO 3?
2 KClO 3 2 KCl + 3 O 2
How many grams of oxygen can be produced from 122.45 g of KClO 3?
1 2 3 4
0% 0% 0% 0%
1 CH 4 + 2 O 2 2 H 2 O + 1 CO 2
How many grams of water are produced from 1. moles of methane?
1 2 3 4
0% 0% 0% 0%
176 g of CO 2 = Number of moles? Molar mass of CO 2 = 44 If you multiply, 176 g x 44 gl^ results in units of g^
2
mole
g
176 g x 44 results in units of you get a unit mess UNITS alert you that you made an error KEEP UNITS WITH NUMBERS!!
mole (^) mole
Find molar mass Divide grams by molar mass
Balance first
H (^) 2 S O 2 SO 2 H 2 O
2 H (^) 2 S 3 O 2 2 SO 2 2 H 2 O
Then determine molar ratios 2 SO 2 to 3 O 2
2 H (^) 2 S 3 O 2 2 SO 2 2 H 2 O
Find molar masses
2 H (^) 2 S 3 O 2 2 SO 2 2 H 2 O
2 H (^) 2 S 3 O 2 2 SO 2 2 H 2 O
2 H (^) 2 S 3 O 2 2 SO 2 2 H 2 O
? O
Is it balanced? Molar ratio 1:1: 4 grams oxygen Grams carbon consumed?
C O 2 CO 2
Grams carbon dioxide produced?
Molar ratio 1:1: 4 grams oxygen 1 mole O 2 = 32 g
C O 2 CO 2
2 0.^1252 32
1 4 molesO g
mole g O
Molar ratio 1:1:
0.125 moles O 2 in 1:1 ratio
C O 2 CO 2
0.125 moles C 0.125 moles CO (^2)
0.125 moles C Grams carbon consumed?
C O 2 CO 2
Collision of molecules required for it to occur Increase concentration Increase temperatureIncrease temperature Catalyst can facilitate reaction