atoms. 2. The correct answer is not given atoms atoms correct

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1 Version 001 HW01-stoichiometry vandenbout (52130) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Mlib points The measurement g could also be written as pg mg. correct kg. 4. None of these g. mg refers to ACAMP FE points The mole concept is important in chemistry because 1. it establishes a standard for reaction stoichiometry. 2. atoms and molecules are very small and themoleconceptallowsustocountatomsand molecules by weighing macroscopic amounts of material. correct 3. it provides a universally accepted standard for mass. 4. it allows us to distinguish between elements and compounds. 5. it explains the properties of gases. The mole concept is important in chemistry because we know that if we weight g of pure copper, then we have about a mole of copper atoms. Mlib points How many atoms of hydrogen are contained in 1 mole of methane (CH 4 )? atoms 2. The correct answer is not given atoms atoms correct 5. 4 atoms n = 1 mol Each methane molecule contains 4 hydrogen atoms. There are Avogadro s number of methane molecules in one mole of methane molecules: n H = 1 mol CH molec CH 4 1 mol CH 4 4 H atoms 1 molec CH 4 = H atoms Counting Hs points Which has the greatest number of hydrogen atoms? 1. 5 g of an unknown compound hydrogen atoms 3.100gofasubstancethatis2%Hbymass g of water g of hydrogen gas correct H atoms is much less than 1 mole of H atoms. 100g of water is 5.56 moles of water which would have moles of H atoms. 5 g of an unknown substance even if it was pure hydrogen could only be 5 moles of H atoms.

2 Version 001 HW01-stoichiometry vandenbout (52130) 2 20gofhydrogengasis10molesofH 2 whichis 20molesofHatoms. 100gofasubstance that is 2% by mass hydrogen has 2 g of Hydrogen which is 2 moles. 20 moles of H atoms is the greatest number of atoms. Mlib points What is the coefficient for H 2 O when the equation?ca(oh) 2 (aq)+?h 3 PO 4 (aq)?ca 3 (PO 4 ) 2 (s)+?h 2 O(l) is balanced using the smallest possible integers? correct A balanced equation has the same number of each kind of atom on each side of the equation. We find the number of each kind of atom using equation coefficients and composition stoichiometry. For example, we find there are 12 H atoms on the product side:?h atoms = 6 H 2 O 2H H 2 O = 12 H The balanced equation is 3Ca(OH) 2 +2H 3 PO 4 Ca 3 (PO 4 ) 2 +6H 2 O, and the coefficient of H 2 O is 6. Mlib points When aluminum metal is heated with manganese oxide, the following reaction occurs. Al+MnO 2 Al 2 O 3 +Mn Balance this equation and indicate the sum of the coefficients for all the species. 1. ten 2. fifteen 3. twelve correct 4. seven A balanced equation has the same number of each kind of atom on both sides of the equation. We find the number of each kind of atom using equation coefficients and composition stoichiometry. For example, we find there are 6 O atoms on the reactant side:? O atoms = 3 MnO 2 2 O 1 MnO 2 = 6 O The balanced equation is 4Al+3MnO 2 2Al 2 O 3 +3Mn, and has 4 Al, 3 Mn and 6 O atoms on each side.? sum coefficients = = 12 Balance Equation points When the equation?pbs+?o 2?PbO+?SO 2 is balanced, the coefficients are 1. 2; 6; 4; ; 2; 1; ; 2; 1; ; 12; 4; ; 3; 2; 2 correct There are 2 oxygens on the left and 3 on the right, so at least six oxygens are needed:?pbs+3o 2 2PbO+2SO 2

3 Version 001 HW01-stoichiometry vandenbout (52130) 3 Now there are 2 each of Pb and S on the right, so the balanced equation is 2PbS+3O 2 2PbO+2SO 2 Balance Equation points Balance the equation?al 2 (SO 4 ) 3 +?NaOH?Al(OH) 3 +?Na 2 SO 4, using the smallest possible integers. What is the sum of the coefficients in the balanced equation? 1. ten 2. eight 3. fourteen 4. twelve correct 5. six The balanced chemical equation is 1Al 2 (SO 4 ) 3 +6NaOH 2Al(OH) 3 +3Na 2 SO 4 which gives 2 Al, 3 SO 4, 6 Na, and 6 OH on both sides of the equation. The sum of coefficients is = 12. Brodbelt points Which one has the greatest number of atoms? 1. All have the same number of atoms moles of argon moles of water moles of CH 4 correct moles of helium For 3.05 moles of water:?atoms = 3.05molH 2 O molec 1mol 3atoms 1molecule = atoms For 3.05 moles of CH 4 :?atoms = 3.05molCH molec 1mol 5atoms 1molecule = atoms For 3.05 moles of helium:? atoms = 3.05 mol He atoms 1 mol = atoms For 3.5 moles of argon:? atoms = 3.05 mol Ar atoms 1 mol = atoms Mlib points If grams of copper (Cu) completely reacts with 25.0 grams of oxygen, how much copper(ii) oxide (CuO) will form from grams of copper and excess oxygen? (Note: CuO is the only product of this reaction.) g g correct g g g m Cu,ini = g m O2 = 25.0 g m Cu,fin = g If 100 g copper and 25 g oxygen react completelywitheachother,theremustbe125gof product formed(law of conservation of mass). This product is CuO.

4 Version 001 HW01-stoichiometry vandenbout (52130) 4 Now we have a ratio: for every 100 g of Cu reacted, 125 g of CuO will be produced (assuming there is enough oxygen). We use this ratio to find the mass of CuO that could beformedfrom140gofcuandexcessoxygen. We set our known ratio (100 g Cu : 125 g CuO) equal to our experimental ratio (140 g Cu : x g CuO) and solve for the unknown: 100 g Cu 140 g Cu = 125 g CuO x (140 g Cu)(125 g CuO) x = 100 g Cu = 175 g CuO Mlib points Consider the reaction 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s). If 12.5 g of iron(iii) oxide (rust) are produced from 8.74 g of iron, how much oxygen gas is needed for this reaction? g g g correct g g m iron = 8.74 g m oxide = 12.5 g The balanced equation for the reaction tells us that 4 mol Fe reacts with 3 mol O 2 to produce 2 mol Fe 2 O 3. We have two possible starting points. We know 12.5 g Fe 2 O 3 was produced and that 8.74 g Fe was present at the start of the reaction. Choosing the 12.5 g of Fe 2 O 3 to start with, firstweconverttomolesusingthemolarmass:? mol Fe 2 O 3 = 12.5 g Fe 2 O 3 1 mol Fe 2O g Fe 2 O 3 = mol Fe 2 O 3 Now we use the mole ratio from the balanced equation to find moles O 2 needed to produce mol Fe 2 O 3.? mol O 2 = mol Fe 2 O 3 3 mol O 2 2 mol Fe 2 O 3 = mol O 2 We convert from moles to grams:? g O 2 = mol O 2 32 g O 2 1 mol O 2 = g O 2 Starting with 8.74 g Fe and following the same steps results in the same numerical answer. Msci points Upon heating, potassium chlorate produces potassium chloride and oxygen: 2KClO 3 2KCl+3O 2. What mass of oxygen (O 2 ) would be produced upon thermal decomposition of 25 g of potassium chlorate (KClO 3 with MW g/mol)? g g correct g g g m KClO3 = 25.0 g MW KClO3 = g/mol The balanced equation for the reaction indicates that 3 mol O 2 are produced for every 2 mol KClO 3 reacted. First we calculate the moles KClO 3 present:? mol KClO 3 = 25 g KClO 3 1 mol KClO g KClO 3 = mol KClO 3

5 Version 001 HW01-stoichiometry vandenbout (52130) 5 Now we use the mole-to-mole ratio from the balanced equation to find the moles O 2 that could be produced from this amount of KClO 3 :? mol O 2 = mol KClO 3 3 mol O 2 2 mol KClO 3 = mol O 2 We convert from moles to grams O 2 :? g O 2 = mol O 2 32 g O 2 1 mol O 2 = 9.8 g O 2 Msci points In the reaction?co+?o 2?CO 2, how much oxygen is required to convert 21 g of CO into CO 2? g g correct g 4. 6 g g g m CO = 21 g The balanced equation for the reaction is 2CO+O 2 2CO 2 The coefficients in this equation indicate that 2mol COareneeded for each molo 2 reacted. First we calculate the moles of CO present: 1 mol CO? mol CO = 21 g CO 28 g CO = 0.75 mol CO Using the mole ratio from the balanced equation, we find the moles O 2 needed to completely react with 0.75 mol CO:? mol O 2 = 0.75 mol CO 1 mol O 2 2 mol CO = mol O 2 We convert from moles to grams O 2 :? g O 2 = mol O 2 32 g O 2 1 mol O 2 = 12 g O 2 Brodbelt points Consider the reaction N 2 +3H 2 2NH 3. How much NH 3 can be produced from the reaction of 74.2 g of N 2 and 14.0 moles of H 2? molecules molecules correct molecules molecules molecules m N2 = 74.2 g n H2 = 14.0 mol First you must determine the limiting reactant:? mol N 2 = 74.2 g N 2 1 mol N 2 28 g N 2 = 2.65 mol N 2 According to balanced equation, we need We have 3 mol H 2 1 mol N mol H mol N 2 = 5.28 mol H 2 1 mol N 2

6 Version 001 HW01-stoichiometry vandenbout (52130) 6 Therefore, H 2 is an excess and N 2 is limiting.? molec NH 3 = 2.65 mol N 2 2 mol NH 3 1 mol N NH 3 molec 1 mol NH 3 molec = molec NH 3 Limit mccord01x hmwk points For the reaction?c 6 H 6 +?O 2?CO 2 +?H 2 O 25.1 grams of C 6 H 6 are allowed to react with grams of O 2. How much CO 2 will be produced by this reaction? Correct answer: grams. m C6H 6 = 25.1 g m O2 = g The balanced equation for the reaction is 2C 6 H 6 +15O 2 12CO 2 +6H 2 O FW of C 6 H 6 is g/mol, giving mol C 6 H 6. FW of O 2 is g/mol, giving mol O 2. FW of CO 2 is g/mol mol C 6 H 6 15 mol O 2 2 mol C 6 H 6= 2.41 mol O2 which is less than what is actually present. ThereforethelimitingreactantmustbeC 6 H mol C 6 H 6 12 mol CO 2 2 mol C 6 H g CO 2 1 mol CO 2 = g CO 2