1Chemical Equation CO(g) + 2H2(g) CH3OH(l) Describes a chemical reactionA + B CA and B = reactantsC = productCO(g) + 2H2(g) CH3OH(l)
2Information Given by Chemical Equations CO(g) + 2H2(g) CH3OH(l)1 molecule of CO + 2 molecules of H2 one molecule of CH3OH10 molecules of CO + 20 molecules of H2 10 molecules of CH3OH1 mole of CO + 2 moles of H2 1 mole of CH3OHDo you notice a relationship pattern?
3StoichiometryProcess of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction.
4Balancing equationsIn a chemical reaction, atoms are neither created nor destroyed.All atoms present in the reactants must be accounted for among the products.Therefore, one needs to balance the chemical equation for the reaction.What goes in must come out, albeit in a different form.
5Rules for balancing equations 1. Balance out metals and polyatomic ions firstConsider polyatomic ions as a single species2. Balance out elements that are found in more than one species lastLeave H for last: Na + H2O NaOH + H23. Don’t get discouraged!Balancing one atom can throw others offBalance the equation afresh4. Do NOT change the nature of the compoundTo balance H2O you cannot make it H2O2Why not?5. Thus, put numbers in front of compounds6. Lowest whole numbersIncorrect: 4A + 4B 4ABCorrect: A + B AB
6Na(s) + H2O(l) NaOH(aq) + H2(g) ExampleBalance:Na(s) + H2O(l) NaOH(aq) + H2(g)Na is balanced, O is balanced, but H is not balance H put 2 in front of NaOHThis gives us 4 H on the products sideAdd 2 in front of water on reactants sideH’s are now balancedSo are O’s (2 on each side)Na is now unbalancedAdd a 2 to Na on reactant sideRecapitulate: check equation2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)BALANCED!
7Balance the following Al(s) + F2(g) AlF3(s) KClO3(s) KCl(s) + O2(g)KI(aq) + Pb(NO3)2(aq) KNO3(aq) + PbI2(s)C2H6(g) + O2(g) CO2(g) + H2O(l)Let’s talk about a neat trick for the above
8Mole-Mole Relationships: Dimensional Analysis Consider the decomposition of water:2H2O(l) 2H2(g) + O2(g)If we decompose 4 mol of H2O, how many moles of O2 do we get?If we decompose 5.8 mol of H2O, how many moles of H2 do we get?
9C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) More practiceCalculate the moles of C3H8 used when 4.30 moles of CO2 are obtained.C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
10Mass Calculations 1. Balance the equation for the reaction. Steps for Calculating the Masses of Reactants and Products in Chemical Reactions1. Balance the equation for the reaction.2. Convert the masses of reactants or products to moles.3. Use the balanced equation to set up the appropriate mole ratio(s).4. Use the mole ratio(s) to calculate the number of moles of the desired reactant or product.5. Convert from moles to mass.
11In other words… gramsA molesA molesA molesB molesB gramsB Do it all as a dimensional analysis problem!See next slide
12Example What mass of 2 should we weigh out to react with 35.0 g Al? 2Al(s) + 32(s) 2Al3(s)gramsA molesA:molesA molesB:molesB gramsB:In one-step:Avoids rounding errors too!
13PracticeWhat mass of CO2 is produced when 96.1 g of C3H8 react with sufficient O2.C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
14Calculations Involving a Limiting Reactant When there is insufficient amount of one of the reactants.Calculate the mass required for the product in question using both reactants.Which ever gives the least amount of product is the limiting reactant.The mass of product obtained from using the limiting reactant is the correct amount.
15ExampleWhat mass of AlI3 will we obtain if we react 35.0 g Al with g I2?2Al(s) + 3I2(s) 2AlI3(s)35.0 g Al x (mol/ g) x (2 mol AlI3/2 mol Al) x ( g/mol) = 529 g AlI3Versus: g I2 x (mol/ g I2) x (2 mol AlI3/3 mol I2) x ( g/mol) = 428 g AlI3 I2 is the limiting reactant.The mass of AlI3 obtained is 428 g.
16PracticeExampleCalculate the mass of lithium nitride formed from 56.0 g of N2 and 56.0 g of Li in the balanced equation:6Li(s) + N2(g) 2Li3N(s)
17More practiceExampleCalculate the mass of aluminum sulfate formed from 50.0 g of Al and 50.0 g of H2SO4 in the balanced equation:2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)
18Percent YieldTheoretical Yield: Amount of product predicted from the amounts of reactants used.Actual Yield: Amount of product actually obtained. Actual yields are determined by doing the reaction__Actual Yield__ X 100% = Percent YieldTheoretical Yield
19TiCl4(g) + O2(g) TiO2(s) + 2Cl2(g) Percent YieldConsider the reaction:TiCl4(g) + O2(g) TiO2(s) + 2Cl2(g)(a) Suppose 6.71 x 103 g of TiCl4 is reacted with 2.45 x 103 g of O2. Calculate the maximum mass of TiO2 that can form.(b) If 2.12 x 103 g was produced in the laboratory, what is the percent yield?