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Chemical Equation

Chemical Equation. Describes a chemical reaction A + B  C A and B = reactants C = product CO (g) + 2H 2(g)  CH 3 OH (l). Information Given by Chemical Equations. CO (g) + 2H 2(g)  CH 3 OH (l) 1 molecule of CO + 2 molecules of H 2  one molecule of CH 3 OH

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Chemical Equation

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  1. Chemical Equation • Describes a chemical reaction • A + B  C • A and B = reactants • C = product CO(g) + 2H2(g)  CH3OH(l)

  2. Information Given by Chemical Equations CO(g) + 2H2(g)  CH3OH(l) • 1 molecule of CO + 2 molecules of H2 one molecule of CH3OH • 10 molecules of CO + 20 molecules of H2 10 molecules of CH3OH • 1 mole of CO + 2 moles of H2 1 mole of CH3OH • Do you notice a relationship pattern?

  3. Stoichiometry • Process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction.

  4. Balancing equations • In 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.

  5. Rules for balancing equations 1. Balance out metals and polyatomic ions first • Consider polyatomic ions as a single species 2. Balance out elements that are found in more than one species last • Leave H for last: Na + H2O  NaOH + H2 3. Don’t get discouraged! • Balancing one atom can throw others off • Balance the equation afresh 4. Do NOT change the nature of the compound • To balance H2O you cannot make it H2O2 • Why not? 5. Thus, put numbers in front of compounds 6. Lowest whole numbers • Incorrect: 4A + 4B  4AB • Correct: A + B  AB

  6. Example • Balance: • 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 NaOH • This gives us 4 H on the products side • Add 2 in front of water on reactants side • H’s are now balanced • So are O’s (2 on each side) • Na is now unbalanced • Add a 2 to Na on reactant side • Recapitulate: check equation • 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) • BALANCED!

  7. Balance 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

  8. Mole-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?

  9. More practice • Calculate the moles of C3H8 used when 4.30 moles of CO2 are obtained. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

  10. Mass Calculations Steps for Calculating the Masses of Reactants and Products in Chemical Reactions 1. 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.

  11. In other words… • gramsA molesA • molesA  molesB • molesB  gramsB • Do it all as a dimensional analysis problem! • See next slide

  12. Example What mass of 2 should we weigh out to react with 35.0 g Al? 2Al(s) + 32(s) 2Al3(s) • gramsA molesA: • molesA  molesB: • molesB  gramsB: • In one-step: • Avoids rounding errors too!

  13. Practice • What mass of CO2 is produced when 96.1 g of C3H8 react with sufficient O2. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

  14. Calculations 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.

  15. Example What mass of AlI3 will we obtain if we react 35.0 g Al with 400.0 g I2? 2Al(s) + 3I2(s) 2AlI3(s) • 35.0 g Al x (mol/26.98154 g) x (2 mol AlI3/2 mol Al) x (407.6950 g/mol) = 529 g AlI3 • Versus: 400.0 g I2 x (mol/253.8090 g I2) x (2 mol AlI3/3 mol I2) x (407.6950 g/mol) = 428 g AlI3 •  I2 is the limiting reactant. • The mass of AlI3 obtained is 428 g.

  16. Practice Example Calculate 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)

  17. More practice Example Calculate 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)

  18. PercentYield • Theoretical 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 Yield Theoretical Yield

  19. Percent Yield Consider 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?

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