* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chem 11 Stoichiometry (mol-mol) Using the formulas we have
Multi-state modeling of biomolecules wikipedia , lookup
Dimensional analysis wikipedia , lookup
Host–guest chemistry wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Electrolysis of water wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Rate equation wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Hypervalent molecule wikipedia , lookup
History of molecular theory wikipedia , lookup
Photosynthetic reaction centre wikipedia , lookup
Determination of equilibrium constants wikipedia , lookup
Transition state theory wikipedia , lookup
Implicit solvation wikipedia , lookup
Self-assembled monolayer wikipedia , lookup
Biochemistry wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Atomic theory wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Size-exclusion chromatography wikipedia , lookup
Computational chemistry wikipedia , lookup
Chem 11 Stoichiometry (mol-mol) Using the formulas we have discussed, we can now get into the real stoichiometry calculations. We will need to look at writing equations and balancing them in order to properly and correctly carry out all stoichiometric calculations. Write the reaction for the formation of Ammonia: N2 + 3H2 2NH3 Proof of the Conservation of Mass: From the balanced equation, we can say that 1 molecule of N2 and 3 molecules of H2 combine to form 2 molecules of NH3 OR we can say that 1 mole of N2 and 3 moles of H2 combine to form 2 moles of NH3. Using the coefficients as moles, we can determine the mass of the each of the reactants present and also for the product formed. Mass of Reactants: m = nxM = (1 mol N2)(28.0134g/mol N2) = 28.0134g N2 m = nxM = (3 mol H2)(2.01588g/mol H2) = 6.04764g H2 Total mass of Reactants = 28.0134g N2 + 6.04764g H2 = 34.06104g Mass of Products: m = nxM = (2 mol NH3)(17.03052g/mol NH3) = 34.06104g Note that the mass of both reactants and products are the same value – Law of Conservation of Mass! Balanced equations allow us to calculate the quantities of reactants and products in a reaction. When you know the quantity (usually grams or moles) of one substance, you can calculate the quantity of any other substance. Recall: STOICHIOMETRY is the calculation of quantities in chemical equations. Consider the information we can get from the following balanced equation: N2(g) + 1 molecule 10 molecules 1 x (6.02 x 1023 molecules) 1 mol N2 3H2(g) 3 molecules 30 molecules 3 x (6.02 x 1023 molecules) 3 mol H2 → 2NH3(g) 2 molecules 20 molecules 2 x (6.02 x 1023 molecules) 2 mol NH3 The most important interpretation of this equation is that 1 mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of ammonia (as we used to determine the mass of reactants and products above). With this information we can relate moles of reactants to moles of products. The coefficients from the balanced equation can be used to write conversion factors. Then the number of moles of a product can be calculated from a given number of moles of reactant. Using other mole-quantity relationships, we can introduce mass, volume, and particles into our calculations (this is what we will be getting to in the next few classes). Example 1: How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? Solution 1: We will work from a known number of moles of N2 to the unknown number of moles of NH3. From the balanced equation just given, we see that 1 mol of N2 produces 2 mol of NH3. This relationship can be written and used as a conversion factor. The conversion factor is called a mole ratio. 1 mol N2_ _2 mol NH3_ OR 2 mol NH3 1 mol N2 To solve the problem, we start with what we’re given, 0.60 mol of N2. Then we multiply it by the form of the mole ratio that allows the units we want to get rid of to cancel. _2 mol NH3_ = 1.2 mol NH3 1 mol N2 (Note that mole ratios from balanced equations are considered to be exact. They do not enter into the determination of significant figures in the answer.) 0.60 mol N2 x Example 2: a) How many moles of aluminum are needed to form 2.3 moles of aluminum oxide? b) Calculate the number of moles of aluminum oxide formed when 17.2 moles of oxygen reacts with aluminum. Solution 2: First we need to write the balanced chemical equation for this reaction: 4 Al + 3 O2 2 Al2O3 Then we use the information given in the question, and our coefficients from our balanced equation, to solve for the moles of the substance we are asked to find. a) nAl = (2.3mol Al2O3)(4 mol Al/2 mol Al2O3) = 4.6mol Al b) n Al2O3 = (17.2mol O2)(2mol Al2O3/3mol O2) = 11.4666mol Al2O3 = 11.5mol Al2O3 Example 3: a) If 3.84 moles of C2H2 burns completely, how many moles of oxygen are needed? b) How many moles of carbon dioxide are produced when 2.47 moles of C2H2 are burned? Solution 3: Again, we write the balanced chemical equation first: 2 C2H2 + 5 O2 4 CO2 + 2 H2O Use our coefficients and the information given in the question to convert to the amount of moles of the substance as asked in the question. a) n O2 = (3.84mol C2H2)(5mol O2/2mol C2H2) = 9.60mol O2 b) n CO2 = (2.47mol C2H2)(4mol CO2/2mol C2H2) = 4.94mol CO2 Complete worksheet – Mole to Mole Practice Problems (#’s 1-6)