Stoichiometry Lesson # 1.

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Presentation transcript:

Stoichiometry Lesson # 1

Stoichiometry: the relationship between the amount of reactants used in a chemical reaction and the amounts of products produced by the reaction

Coefficients in a Reaction Equation: 2Mg + O2 → 2MgO the coefficient 2, means two magnesium atoms react with one oxygen molecule to produce two molecules of magnesium oxide if you double the amounts of Mg and O2, what will happen to the amount of MgO??? if you use 10 times the amount of reactants, what will happen to the amount of MgO???

*the balanced equation describes the ratio in which the substances combine In terms of MOLES: (remember 1 mole =6.02x1023 molecules) 2(6.02 x 1023)Mg + (6.02 x 1023)O2 → 2(6.02 x 1023)MgO 2 mol Mg + 1 mol O2 → 2 mol MgO The ratio Mg = 2 2 mol Mg O2 1 1 mol O2

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.

So, how do we relate these amounts of moles to determine the relationship. THE MOLE BRIDGE!! It is the MOLE that allows us to compare the amounts (grams) or Atoms required or produced in the reaction.

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 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 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 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 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 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 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 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 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 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 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 Volume of a Gas at STP Lesson # 2

The molar volume of any gas at STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP.

The molar volume of any gas at STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 mole memorize! Calculate the volume of 10.0 lbs of CO2 at STP. 10.0 lbs

The molar volume of any gas at STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 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 at STP standard temperature (25 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 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 at STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 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 at STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 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 at STP standard temperature (0 0C) and pressure (101kPa) is 22.4 L. Molar Volume is 22.4 L 1 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 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 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 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 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 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 diatomic 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 = grams moles = 10.62 g = 28.0 g/mole 0.37946 moles The gas is N2