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CHEM I Chapter 9 Chemical Quantities Review Chapter 8 Converting Grams to moles A sample of hydrogen has a mass of .500 grams how many moles does this sample contain? 0.500 grams H x 1 mol H = 0.496 mol H in sample 1.008 g H Converting moles to grams 9.1 Information given by chemical equations -Chemical changes: rearrangements of atom groupings that are described by chemical equations Consider the balanced chemical equation CO + 2H2 CH3OH -This is the smallest possible combination of atoms that still represents the reaction -But the equation is still balanced even if we multiply this by 12 12CO + 24H2 12 CH3OH -We could even use other units to balance the equation 1dozen CO + 2Dozen H2 1 dozen CH3OH This still would factor down to CO + 2H2 CH3OH -We could also use a really large number like 6.02 x 1023 6.02 x 1023CO + 2(6.02 x 1023)H2 6.02 x 1023CH3OH -But what do we call this number 6.02 x 1023 : mole So this equation would also be written as 1 mol CO + 2 mol H2 1 mol CH3OH All the balanced chemical equations that we have done can also be expressed in this way C3H8 + O2 CO2 + H2O Balance this! C3H8 + 5O2 3CO2 + 4H2O Interpret in terms of molecules 1 molecule of C3H8 reacts with 5 molecules of O2 to give 3 molecules of CO2 plus 4 molecules of H2O Interpret in terms of moles 9.2 1 mol of C3H8 reacts with 5 mol of O2 to give 3 mol of CO2 plus 4 mol of H 2O Mole-Mole Relationships -We can use the chemical equation to predict how much product will come from a given amount of product. For example: take this balanced chem.. equation for the decomposition of water 2H2O 2H2 + O2 This would read as 2 mol of water yields 2 mol of hydrogen gas and 1 mol of oxygen gas What if we start with 4 mole of water, How many moles of each product would we get? Think of it as a ratio 2 H2 = x H2 2H2O 4H2O x = 4 mol H2 1O2 2H2O = x O2 4H2O x = 2 mol O2 What have we done to the equation? 2[2H2O 2H2 + O2] -we multiplied the entire equation by two 4H2O 4H2 + 2O2 What if we were to start with 5.8 mol of water, How many moles of products would this yield? 2 ways to do this 1 the ratio way we did before 2 H2 = x H2 x = 5.8 mol H2 2H2O 5.8H2O 1O2 = x O2 2H2O 5.8H2O x = 2.9 mol O2 Or to make the mole ratios: conversion factors based on the balanced chemical equation. Before you make the ratio 2H2O 2H2 + O2 get H2O to 1 by dividing by 2 H2O H2 + ½ O2 Now multiply by 5.8 5.8H2O 5.8H2 + 2.9 O2 -this second way is the way your book does this problems or use it as a conversion factor ex. 5.8 mol H2O x 1 mol O2 = 2.9 mol O2 2 mol H2O -this is how I like the problem done Ex. 9.3 Mole Ratio in Calculations page263 C3H8 + 5O2 3 CO2 + 4H2O Calculate the number of moles of oxygen required to react with exactly 4.30 mol of propane. C3H8 in the reaction 4.30 mol C3H8 x 5 mol O2 = 21.5 mol O2 1 mol C3H8 How many CO2 would be produced, how many H2O? 4.30 mol C3H8 x 3 mol CO2 = 12.9 mol CO2 1mol C3H8 4.30 mol C3H8 x = 17.2 mol H2O 4 mol H2O 1 mol C3H8 9.3 Mass Calculations -review for converting moles to masses given the equations 2Al + 3I2 2 AlI3 We have 35 g of Aluminum What mass of I2 should we weigh out to react exactly with this amount of aluminum? 2 mol of Al requires 3 mole of I2 our mole ratio is 3 mol I2 2 mol Al Since the ratio is in moles our amount of Al must also be in moles Grams of Al x 1 mol Al = moles of Al Molar mass of AL 35.0 g Al x 1 mol Al x 3 mol I2 = 1.95 mol I2 26.98 g Al 2 mol Al Now we know the number of moles I2 required how many grams is that? 1.95 mol I2 x 253.8 g I2 = 495 g I2 1 mol I2 Ex. 9.4 page 265 Given C3H8 + 5O2 3CO2 + 4H2O (make sure its balanced) What mass of oxygen will be required to react with exactly 96.1 g of Propane? 1st = change to moles 2nd = use mole ratio to convert to other chemical 3rd = use molar mass to convert from moles to grams 96.1 g C3H8 x 1 mol C3H8 x 44.09 g C3H8 5 mol O2 1 mol C3H8 = 10.9 mol O2 10.9 mol O2 x 32.0 g O2 = 349 g O2 1 mol O2 Steps for calculating the masses of reactants and products of chemical reactions 1. Balance the equation 2. convert the masses to moles 3. use mole ratio from balanced equation to convert from moles of one chemical to another 4. Convert from moles to mass This process of using the chemical equation to calculate the relative masses of reactants and products involved in a reaction is called stoichiometry -the balanced equation describes the Stoichiometry of the reaction Using Stoichiometry Ex. 9.5 page 268 When solid lithium hydroxide and carbon dioxide gas react the products are solid lithium carbonate and liquid water. What mass of carbon dioxide gas can 1000 grams of lithium hydroxide absorb? 1st write the equation: LiOH (s) + CO2 (g) Li2CO3 (s) + H2O(l) nd 2 balance the equation: 2LiOH (s) + CO2 (g) Li2CO3 (s) + H2O(l) 3rd start the problems 1st convert to moles: 1000 g LiOH x 1 mol LiOH/23.95 g LiOH = 41.8 mol LiOH nd 2 use mole ratio to convert between chemicals 41.8 mol LiOH x 1 mol CO2/ 2 mol LiOH = 20.9 mol CO2 rd 3 convert to grams 20.9 mol CO2 x 44.01 g CO2/ 1 mol CO2 = 920 g CO2 9.4 Calculations involving Limiting Reactant -In a chemical rxn: the reaction can only proceed as long as enough of all the parts are present -for example: if you were making a bologna sandwich you have 4 pieces of bread 2 pieces of cheese 1 slice of bologna -How many complete sandwiches can you make? 1 We only have 1 slice of bologna Ex. CH4 + H2O 3H2 + CO What mass of water is required to react exactly with 249 grams of methane? 249 g x 1mol/16.04 g = 15.5 mol CH4 15.5 mol x 1 mol H2O/1 mol CH4 = 15.5 mol H2O 15.5 mol x 18.02 g H2O/ 1 mol = 279 g H2O -This means that we need exactly 279 grams of water before all the methane is used up. If we only had 200 grams of water for 249 grams of methane? -not all of the methane would be consumed -Water is limiting the amount of products that can be made -Limiting reactant/reagent: a reactant that is consumed first in the reaction that limits the amount of products that can be made If we had 300 grams of water for 249 grams of methane? -We would have excess water -Water would be in excess. -methane would be the limiting reactant -Always check to see which reactant is limiting when you are given 2 or more initial amounts -Using the mole ratio determine which reactant is limiting -then use the limiting reactant to compute the amt of product that is produced Ex. 9.8 page 277 Nitrogen gas can be prepared by passing ammonia gas over solid copper II oxide at high temps. The other products of the reaction are solid copper and water vapor. How many grams of N2 are formed when 18.1 grams of NH3 are reacted with 90.4 g CuO? 2NH3 + 3CuO N2 + 3Cu + 3H2O 18.1 g NH3 x 1 mol / 17.03 g = 1.06 mol NH3 90.4 g CuO x 1 mol / 79.55 g CuO = 1.14 mol CuO Determine limiting reactant: use the mole ration between the 2 reactants 1.06 mol NH3 x 3 mol CuO/2 mol NH3 = 1.59 mol CuO We have less mol CuO than we need if 1.59 mol CuO is required to react with 1.06 mol NH3 then the CuO will run out before the NH3 and the CuO is the limiting reactant. 1.14 mol CuO x 1 mol N2/ 3 mol CuO = 0.380 mol N2 0.380 mol x 28.02 g/1 mol = 10.6 g N2 9.5 Percent Yield: -Theoretical yield: the amount of product that will be made calculated from Stoichiometry -the amt of reactants predicts the amount of product made -IT shows the maximum amount of product that can be made from the reactants -Actual Yield: the amount of product actually obtained from a reaction - usually done experimentally -Percent yield: comparison of the actual yield to the theoretical yield - Actual yield/ theoretical yield x 100% = percent yield Ex. If a reaction produces 6.63 g of N2 but the predicted amount is 10.6 g of N2 what is the percent yield? 6.63 g / 10.6 g x 100% = 62.5%