A certain buffer is made by dissolving NaHCO3 and Na2CO3

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