stoichiometry lesson # 1
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Stoichiometry Lesson # 1. Balanced chemical equations can be used to relate the amounts of reactants and products in chemical reactions. 2 Fe 2 O 3 + 3 C→ 4 Fe+ 3 CO 2 - PowerPoint PPT PresentationTRANSCRIPT
StoichiometryLesson # 1
Balanced chemical equations can be used to relate the amounts of reactants and products in chemical reactions.
2Fe2O3 + 3C → 4Fe + 3CO2
The coefficients mean that 2 moles of Fe2O3 react with 3 moles of C to
produce 4 moles of Fe and 3 moles of CO2.
The coefficients are a chemical recipe that describes the exact amounts of reactants required to make exact amounts of products in moles.
1. How many grams of Fe2O3 are required to produce 105 g of
Fe?
2Fe2O3 + 3C → 4Fe + 3CO2
?g 105 g
1. How many grams Fe2O3 of are required to produce 105 g of
Fe?
2Fe2O3 + 3C → 4Fe + 3CO2
?g 105 g
105 g Fe
1. How many grams Fe2O3 of are required to produce 105 g of
Fe?
2Fe2O3 + 3C → 4Fe + 3CO2
?g 105 g
105 g Fe x 1 mole 55.8 g
1. How many grams Fe2O3 of are required to produce 105 g of
Fe?
2Fe2O3 + 3C → 4Fe + 3CO2
?g 105 g
105 g Fe x 1 mole x 2 mole Fe2O3
55.8 g 4 mole Fe
1. How many grams Fe2O3 of are required to produce 105 g of
Fe?
2Fe2O3 + 3C → 4Fe + 3CO2
?g 105 g
105 g Fe x 1 mole x 2 mole Fe2O3 x 159.6 g
55.8 g 4 mole Fe 1 mole
1. How many grams Fe2O3 of are required to produce 105 g of
Fe?
2Fe2O3 + 3C → 4Fe + 3CO2
?g 105 g
105 g Fe x 1 mole x 2 mole Fe2O3 x 159.6 g = 150. g
55.8 g 4 mole Fe 1 mole
2. How many grams of C are require to consume 155 g of Fe2O3?
2Fe2O3 + 3C → 4Fe + 3CO2
155g ? g
2. How many grams of C are require to consume 155 g of Fe2O3?
2Fe2O3 + 3C → 4Fe + 3CO2
155g ? g
155 g Fe2O3
2. How many grams of C are require to consume 155 g of Fe2O3?
2Fe2O3 + 3C → 4Fe + 3CO2
155g ? g
155 g Fe2O3 x 1 mole
159.6 g
2. How many grams of C are require to consume 155 g of Fe2O3?
2Fe2O3 + 3C → 4Fe + 3CO2
155g ? g
155 g Fe2O3 x 1 mole x 3 mole C
159.6 g 2 mole Fe2O3
2. How many grams of C are require to consume 155 g of Fe2O3?
2Fe2O3 + 3C → 4Fe + 3CO2
155g ? g
155 g Fe2O3 x 1 mole x 3 mole C x 12.0 g
159.6 g 2 mole Fe2O3 1 mole
2. How many grams of C are require to consume 155 g of Fe2O3?
2Fe2O3 + 3C → 4Fe + 3CO2
155g ? g
155 g Fe2O3 x 1 mole x 3 mole C x 12.0 g = 17.5 g
159.6 g 2 mole Fe2O3 1 mole
3. How many grams of Al2(CO3)3 are produced by the complete
reaction of 452 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3
452 g ? g
3. How many grams of Al2(CO3)3 are produced by the complete
reaction of 452 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 2
452 g ? g
452 g Al(NO3)3
3. How many grams of Al2(CO3)3 are produced by the complete
reaction of 452 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 2
452 g ? g
452 g Al(NO3)3 x 1 mole
213.0 g
3. How many grams of Al2(CO3)3 are produced by the complete
reaction of 452 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 2
452 g ? g
452 g Al(NO3)3 x 1 mole x 1 mole Al2(CO3)3
213.0 g 2 mole Al(NO3)3
3. How many grams of Al2(CO3)3 are produced by the complete
reaction of 452 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3 2
452 g ? g
452 g Al(NO3)3 x 1 mole x 1 mole Al2(CO3)3 x 234.0 g = 248 g
213.0 g 2 mole Al(NO3)3 1 mole
4. How many moles of Na2CO3 are required to completely
consume 152 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3
152 g ? moles
4. How many moles of Na2CO3 are required to completely
consume 152 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3
152 g ? moles
152 g Al(NO3)3
4. How many moles of Na2CO3 are required to completely
consume 152 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3
152 g ? moles
152 g Al(NO3)3 x 1 mole
213.0 g
4. How many moles of Na2CO3 are required to completely
consume 152 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3
152 g ? moles
152 g Al(NO3)3 x 1 mole x 3 mole Na2CO3
213.0 g 2 mole Al(NO3)3
4. How many moles of Na2CO3 are required to completely
consume 152 g of Al(NO3)3?
2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3
152 g ? moles
152 g Al(NO3)3 x 1 mole x 3 mole Na2CO3 = 1.07 moles
213.0 g 2 mole Al(NO3)3
Molar Volumeof a Gas at
STPLesson # 2
The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L.
Molar Volume is 22.4 L1 mole memorize!
Calculate the volume of 10.0 lbs of CO2 at STP.
The molar volume of any gas a STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L.
Molar Volume is 22.4 L1 mole memorize!
Calculate the volume of 10.0 lbs of CO2 at STP.
10.0 lbs
The molar volume of any gas a STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L.
Molar Volume is 22.4 L1 mole memorize!
Calculate the volume of 10.0 lbs of CO2 at STP.
10.0 lbs x 1.00 kg 2.21 lbs
The molar volume of any gas a STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L.
Molar Volume is 22.4 L1 mole memorize!
Calculate the volume of 10.0 lbs of CO2 at STP.
10.0 lbs x 1.00 kg x 1000 g 2.21 lbs 1 kg
The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L.
Molar Volume is 22.4 L1 mole memorize!
Calculate the volume of 10.0 lbs of CO2 at STP.
10.0 lbs x 1.00 kg x 1000 g x 1 mole 2.21 lbs 1 kg 44.0 g
The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L.
Molar Volume is 22.4 L1 mole memorize!
Calculate the volume of 10.0 lbs of CO2 at STP.
10.0 lbs x 1.00 kg x 1000 g x 1 mole x 22.4 L 2.21 lbs 1 kg 44.0 g 1 mole
The molar volume of any gas a STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L.
Molar Volume is 22.4 L1 mole memorize!
1. Calculate the volume of 10.0 lbs of CO2 at STP.
10.0 lbs x 1.00 kg x 1000 g x 1 mole x 22.4 L = 2.30 x 103 L 2.21 lbs 1 kg 44.0 g 1 mole
2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO.
2HgO → 2Hg + O2
12.5 g ? L
12.5 g HgO
2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO.
2HgO → 2Hg + O2
12.5 g ? L
12.5 g HgO x 1mole 216.6g
2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO.
2HgO → 2Hg + O2
12.5 g ? L
12.5 g HgO x 1mole x 1 mole O2 216.6g 2 mole HgO
2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO.
2HgO → 2Hg + O2
12.5 g ? L
12.5 g HgO x 1mole x 1 mole O2 x 22.4 L 216.6g 2 mole HgO 1 mole
2. Calculate the volume of O2 gas produced at STP for the complete decomposition of 12.5 g HgO.
2HgO → 2Hg + O2
12.5 g ? L
12.5 g HgO x 1mole x 1 mole O2 x 22.4 L = 0.646 L 216.6g 2 mole HgO 1 mole
3. 10.62 g of a common gas occupies 8.50 L at STP. Calculate the molar mass of the gas and determine the gas.
8.50 L x 1 mole = 0.37946 moles 22.4 L
Molar Mass = gramsmoles
= 10.62 g = 28.0 g/mole0.37946 moles
The gas is N2