A certain buffer is made by dissolving NaHCO3 and Na2CO3 in some water. Write equations to show how this buffer neutralizes added H and OH.

CHM2045 Gower 1-21-16 Two Equation Problem What maximum number of moles of D can be prepared if 2.00 mol of A are combined with 2.00 mol of B and equation 1 is 75% efficient and equation 2 is 50% efficient Equation 1: 2A + B → 4C Equation 2: 3C → 2D + E We must go through several steps using the given values before we can actually figure out how much D can be prepared Start by figuring out if A or B is the limiting reactant in the reaction using the given values o 2 mol A x (4 mol C/2 mol A) = 4 mol C o 2 mol B x (4 mol C/1 mol B) = 8 mol C A is the limiting reactant and B is the excess reactant This simplifies the problem a bit, as equation 1 already begins with 2 mol of A and we can tell that 2 mol of A will make 4 mol of C However, you must remember we are told equation 1 is only 75% efficient, so the actual yield of equation 1 will be 3 mol C [(4 mol C x 0.75)= 3 mol C] Now we can use the amount of C to find the maximum amount of D we can have prepared o 3 mol C x (2 mol D/3 mol C) = 2 mol D This works out well like with equation 1, and we know that 2 mol of D can be made with 3 mol of C However, we must remember that we are told that equation 2 is only 50% efficient, so we must take that into account before we submit a final answer o 2 mol D x 0.50 = 1 mol D The final answer will be 1 mol of D can be prepared from 2 mol of A Solution Stoichiometry Concentration of a solution can be found by using amount of solute present in the given quantity of the solvent or solution o M = molarity = [(moles of solute/liters of solution) x 100] = (mol/L) What mass of KI is needed to make 500mL of a 2.8 M KI solution Begin by converting mL to L o 500mL =0.500L Set unknown number of moles = X 2.8 M = (X mol/0.500L) 2.8 M x 0.500L = X mol 1.4 = X mol You now know the moles of KI needed to make 0.500L of a 2.8 M solution and you can use stoichiometry to convert to grams and get the mass of KI needed o 1.4 mol KI x (166g KI/1 mol KI) = 232g of KI Dissolving Solutions One mole of an ionic solid(nonmetal + metal compound) yields two or more moles of ions when dissolved in water o Solution are electrolytic + - o KBr(s) → K (aq) + Br (aq) 1 mole of ionic sold = 1 mole of cation + 1 mole of anion 2 moles of product o Na 2O (3) → 2Na (aq) + CO (a3)- 1 mole of ionic solid = 2 moles of cation + 1 mole of anion 3 moles of product If you dissolve 5 mol of (NH ) 4 2in 4 O, h2w many mol of NH 4+ and SO 42-are in the resulting solution Try writing this equation yourself—the answer is 10 mol NH 4+and 5 mol SO 42- What is the mass you would need to prepare 250mL of a 0.30 M K CrO solution2 4 0.30M x (X mol/0.250L) = 0.075 mol K CrO x (294.24/1 mol) = 15g K CrO 2 4 IMPORTANT: WILL BE ON FIRST EXAM 75mL of a 2.0 M solution is combined with 125mL of a 4.0 M Na SO . What 2s t4e resulting molarity of the Na ions in the mixture This important to understand, so I will walk you through it, but you must do the math yourself Begin by finding the amount of moles of the solvent in each solution Convert from moles of solution to moles of Na for each amount of moles Add together the moles of Na and divide sum by 0.200L(125mL + 75mL) You should end up with 5.8 M of Na + Dilution Procedure for preparing less concentrated solution from a more concentrated solution o REMEMBER this equation- MV = M Vi i f f o M=iinitial molarity o Vi initial volume o M = final molarity f o Vf= final volume CHM2045 Gower 1-20-16 How To Read Chemical Equations → 2Mg+O 2 2MgO o 2 moles of magnesium + 1 mole of oxygen becomes 2 moles of magnesium oxide o REMEMBER: mass of the reactants must equal the mass of the products o The chemical equation must be balanced Meaning you must have the same amount of moles of an element on the reactant side as you do on the product side o Hint: combustion reactions, like this one, usually create a product that ends in “-oxide” → Try to balance C 2 6 O 2 2CO +23H O 2 Note first that the only element needing to be balanced is oxygen, as carbon and hydrogen already have been balance on both sides of the equation Count the number of oxygen moles on the product side of the equation [(2 x O2)+(3 x O)] = 7 oxygen moles on product side o C 2 6 (7/2)O 2 → 2CO +23H O 2 Simply by multiplying all coefficients by 2 o 2C 2 6 7O 2 → 4CO 2 6H O 2 o That is your balanced equation, with 4 moles of carbon on both sides, 12 moles of hydrogen on both sides and 14 moles of oxygen on both sides Methanol burns in air. If 209 g of methanol are used in the reaction, what mass of H O is produced 2 Begin with the balanced equation 2CH OH3+ 3O 2 → 2CO 2 4H O 2 You must use stoichiometry to go from grams of methanol to grams of water 209g CH O3 x (1 mol CH OH332.042g CH OH) x3(4 mol H O/2 mo2 CH OH) x (13.016g H 2/1 mol H O2 = 235g of H O 2 REMEMBER: When doing these kinds of problems, you are operating under the assumption that there is enough reactant to get a full product--that there is enough O 2or all the methanol to react fully. Limiting Reactants You have a limiting reactant when there is not enough of one reactant to let all of the other reactant get fully used up in the reaction Usually this is a reactant that needs a coefficient in front of it to balance the equation o 2NO + O 2 → 2NO 2 o In this case we needed 2 moles of NO to react with 1 mole of O , 2 therefore NO is the limiting reactant, and O is th2 excess reactant. 124g of Al reacted with 601g of Fe O . W2a3 is the mass of the product Al O 2 3 Begin by writing out the balanced equation to get the molar ratios o 2Al + Fe O2 3 → Al2O 3 2Fe Use stoichiometry to find the moles of aluminum oxide created when 124g Al reacts fully and when 601g of Fe O re2ct3 fully 124g Al x (1 mol Al/26.98g Al) x (1 mol Al O /2 2o3 Al) =2.3 mol Al O 2 3 601g Fe O2x 31 mol Fe O /129.3g Fe O ) x (2 3ol Al O /1 mol2Fe3O ) =3.76 2ol3Al O 2 3 Based on the above equations, we can tell that Al is the limiting reactant in this problem and iron (III) oxide was the excess reactant. We also now aware that the most Al O that2ca3 be produced is 2.3 mol which converts to 235g of Al O 2 3 Note- to find amount of excess reactant is leftover, just convert from moles of limiting reactant to moles of excess reactant and subtract that from the moles of excess reactant available o 124g Al x (1 mol Al/26.98g Al) x (1 mol Fe O /2 2ol3Al) =2.3 mol Fe O 2 3 o 3.76 mol Fe O 2 3.3 mol Fe O = 1253mol Fe O leftove2 3 Reaction Yield Theoretical Yield: what you calculate to be the mass of the product in a reaction Actual Yield: the actual amount of product created by the reaction Percent Yield =(Actual yield/Theoretical yield) x 100 THERE IS ALWAYS A QUESTION ON AT LEAST THE FIRST EXAM AND THE FINAL ABOUT PERCENT YIELD. Double Unknown questions are the hardest you will encounter on this exam. A 29.20g mixture of benzene (C H ) a6d 6oluene (C H ) is c7mb8sted in excess O a2d 98.09g of CO is yie2ded. What were the individual masses of benzene and toluene in the original mixture Begin by converting carbon dioxide to moles o 98.09g CO x 21 mol CO /44.02g CO ) = 2.232mol CO 2 You know from 1 mole of C H yo6 w6ll get at least 6 moles of CO (to balanc2 the equation) For the same reason you know you will have at least 7 moles of CO for each 2 mole of C H7 8 Set x= grams of benzene Set moles of CO pro2uced to the stoichiometry you will need to get grams of C 6 6nd C H 7 8 o 2.23=[(x g C H ) x (1 mol C H /78.11g C H ) x (6 mol CO /1 mol C H ) 6 6 6 6 6 6 2 6 6 Use algebra to solve for x and you will have your grams of benzene and toluene o 12.05g benzene and 17.15g of toluene CHM2045 Gower 1-21-16 Two Equation Problem What maximum number of moles of D can be prepared if 2.00 mol of A are combined with 2.00 mol of B and equation 1 is 75% efficient and equation 2 is 50% efficient Equation 1: 2A + B → 4C Equation 2: 3C → 2D + E We must go through several steps using the given values before we can actually figure out how much D can be prepared Start by figuring out if A or B is the limiting reactant in the reaction using the given values o 2 mol A x (4 mol C/2 mol A) = 4 mol C o 2 mol B x (4 mol C/1 mol B) = 8 mol C A is the limiting reactant and B is the excess reactant This simplifies the problem a bit, as equation 1 already begins with 2 mol of A and we can tell that 2 mol of A will make 4 mol of C However, you must remember we are told equation 1 is only 75% efficient, so the actual yield of equation 1 will be 3 mol C [(4 mol C x 0.75)= 3 mol C] Now we can use the amount of C to find the maximum amount of D we can have prepared o 3 mol C x (2 mol D/3 mol C) = 2 mol D This works out well like with equation 1, and we know that 2 mol of D can be made with 3 mol of C However, we must remember that we are told that equation 2 is only 50% efficient, so we must take that into account before we submit a final answer o 2 mol D x 0.50 = 1 mol D The final answer will be 1 mol of D can be prepared from 2 mol of A Solution Stoichiometry Concentration of a solution can be found by using amount of solute present in the given quantity of the solvent or solution o M = molarity = [(moles of solute/liters of solution) x 100] = (mol/L) What mass of KI is needed to make 500mL of a 2.8 M KI solution Begin by converting mL to L o 500mL =0.500L Set unknown number of moles = X 2.8 M = (X mol/0.500L) 2.8 M x 0.500L = X mol 1.4 = X mol You now know the moles of KI needed to make 0.500L of a 2.8 M solution and you can use stoichiometry to convert to grams and get the mass of KI needed o 1.4 mol KI x (166g KI/1 mol KI) = 232g of KI Dissolving Solutions One mole of an ionic solid(nonmetal + metal compound) yields two or more moles of ions when dissolved in water o Solution are electrolytic + - o KBr(s) → K (aq) + Br (aq) 1 mole of ionic sold = 1 mole of cation + 1 mole of anion 2 moles of product o Na 2O (3) → 2Na (aq) + CO (a3)- 1 mole of ionic solid = 2 moles of cation + 1 mole of anion 3 moles of product If you dissolve 5 mol of (NH ) 4 2in 4 O, h2w many mol of NH 4+ and SO 42-are in the resulting solution Try writing this equation yourself—the answer is 10 mol NH 4+and 5 mol SO 42- What is the mass you would need to prepare 250mL of a 0.30 M K CrO solution2 4 0.30M x (X mol/0.250L) = 0.075 mol K CrO x (294.24/1 mol) = 15g K CrO 2 4 IMPORTANT: WILL BE ON FIRST EXAM 75mL of a 2.0 M solution is combined with 125mL of a 4.0 M Na SO . What 2s t4e resulting molarity of the Na ions in the mixture This important to understand, so I will walk you through it, but you must do the math yourself Begin by finding the amount of moles of the solvent in each solution Convert from moles of solution to moles of Na for each amount of moles Add together the moles of Na and divide sum by 0.200L(125mL + 75mL) You should end up with 5.8 M of Na + Dilution Procedure for preparing less concentrated solution from a more concentrated solution o REMEMBER this equation- MV = M Vi i f f o M=iinitial molarity o Vi initial volume o M = final molarity f o Vf= final volume CHM2045 Gower 1-20-16 How To Read Chemical Equations → 2Mg+O 2 2MgO o 2 moles of magnesium + 1 mole of oxygen becomes 2 moles of magnesium oxide o REMEMBER: mass of the reactants must equal the mass of the products o The chemical equation must be balanced Meaning you must have the same amount of moles of an element on the reactant side as you do on the product side o Hint: combustion reactions, like this one, usually create a product that ends in “-oxide” → Try to balance C 2 6 O 2 2CO +23H O 2 Note first that the only element needing to be balanced is oxygen, as carbon and hydrogen already have been balance on both sides of the equation Count the number of oxygen moles on the product side of the equation [(2 x O2)+(3 x O)] = 7 oxygen moles on product side o C 2 6 (7/2)O 2 → 2CO +23H O 2 Simply by multiplying all coefficients by 2 o 2C 2 6 7O 2 → 4CO 2 6H O 2 o That is your balanced equation, with 4 moles of carbon on both sides, 12 moles of hydrogen on both sides and 14 moles of oxygen on both sides Methanol burns in air. If 209 g of methanol are used in the reaction, what mass of H O is produced 2 Begin with the balanced equation 2CH OH3+ 3O 2 → 2CO 2 4H O 2 You must use stoichiometry to go from grams of methanol to grams of water 209g CH O3 x (1 mol CH OH332.042g CH OH) x3(4 mol H O/2 mo2 CH OH) x (13.016g H 2/1 mol H O2 = 235g of H O 2 REMEMBER: When doing these kinds of problems, you are operating under the assumption that there is enough reactant to get a full product--that there is enough O 2or all the methanol to react fully. Limiting Reactants You have a limiting reactant when there is not enough of one reactant to let all of the other reactant get fully used up in the reaction Usually this is a reactant that needs a coefficient in front of it to balance the equation o 2NO + O 2 → 2NO 2 o In this case we needed 2 moles of NO to react with 1 mole of O , 2 therefore NO is the limiting reactant, and O is th2 excess reactant. 124g of Al reacted with 601g of Fe O . W2a3 is the mass of the product Al O 2 3 Begin by writing out the balanced equation to get the molar ratios o 2Al + Fe O2 3 → Al2O 3 2Fe Use stoichiometry to find the moles of aluminum oxide created when 124g Al reacts fully and when 601g of Fe O re2ct3 fully 124g Al x (1 mol Al/26.98g Al) x (1 mol Al O /2 2o3 Al) =2.3 mol Al O 2 3 601g Fe O2x 31 mol Fe O /129.3g Fe O ) x (2 3ol Al O /1 mol2Fe3O ) =3.76 2ol3Al O 2 3 Based on the above equations, we can tell that Al is the limiting reactant in this problem and iron (III) oxide was the excess reactant. We also now aware that the most Al O that2ca3 be produced is 2.3 mol which converts to 235g of Al O 2 3 Note- to find amount of excess reactant is leftover, just convert from moles of limiting reactant to moles of excess reactant and subtract that from the moles of excess reactant available o 124g Al x (1 mol Al/26.98g Al) x (1 mol Fe O /2 2ol3Al) =2.3 mol Fe O 2 3 o 3.76 mol Fe O 2 3.3 mol Fe O = 1253mol Fe O leftove2 3 Reaction Yield Theoretical Yield: what you calculate to be the mass of the product in a reaction Actual Yield: the actual amount of product created by the reaction Percent Yield =(Actual yield/Theoretical yield) x 100 THERE IS ALWAYS A QUESTION ON AT LEAST THE FIRST EXAM AND THE FINAL ABOUT PERCENT YIELD. Double Unknown questions are the hardest you will encounter on this exam. A 29.20g mixture of benzene (C H ) a6d 6oluene (C H ) is c7mb8sted in excess O a2d 98.09g of CO is yie2ded. What were the individual masses of benzene and toluene in the original mixture Begin by converting carbon dioxide to moles o 98.09g CO x 21 mol CO /44.02g CO ) = 2.232mol CO 2 You know from 1 mole of C H yo6 w6ll get at least 6 moles of CO (to balanc2 the equation) For the same reason you know you will have at least 7 moles of CO for each 2 mole of C H7 8 Set x= grams of benzene Set moles of CO pro2uced to the stoichiometry you will need to get grams of C 6 6nd C H 7 8 o 2.23=[(x g C H ) x (1 mol C H /78.11g C H ) x (6 mol CO /1 mol C H ) 6 6 6 6 6 6 2 6 6 Use algebra to solve for x and you will have your grams of benzene and toluene o 12.05g benzene and 17.15g of toluene