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Stoichiometry Monday, January 29, 2024 6:44 PM CHEM 107 Stoichio... Chemistry Learning Laboratory CHEM 107: Stoichiometry, Limiting Reagents, Percent Yield, and Solution Stoichiometry .CE; MASTERY MAP: Track your progress Learning Outcome TypicaI.Test | can do it on | can do it after oonlf:rsnegtglher Questions my own getting help etting hel Convert grams or moles of a reactant into either grams or moles of a 1,2,3,4 product. Identify the limiting reagent in a reaction. 5678910 Determine grams or moles of a substance formed when one of the 5,6,7,8,9, 10 reactants is limiting. Determine th_e percenl_ yield for a 11,12,13 chemical reaction. Determine the concentration or volume of a product that forms in 14, 15, 16 solution. Consider the following reaction when answering questions 1 and 2. 4NH;3(g) + 50,(g) — 4N0(g) + 6H,0(g) 1. How many moles of water, H;O, are formed when 2.5 moles of ammonia, NH3, react with excess oxygen? First, determine the conversion factor that you will use to convert from moles of ammonia to moles of water. 6 mol H,0 q mol NH3 Use the conversion factor above to calculate the number of moles of water formed. Round your answer to the correct number of significant figures. 15V, , fi%)= @5 Mles of H@ 2. Apply what you have leamed above to answer the following question. How many moles of oxygen, Oz, must have reacted with excess ammonia to produce 6.3 moles of nitrogen monoxide, NO. Round your answer to the correct number of significant figures. lIN\WB 1875 mol D Consider the following reaction when answering questions 3 and 4. C3Hg(g) + 50,(g) - 3C0,(g) + 4H,0(g) 3. How many grams of water, H,O, form when 50.0 g of propane, C3Hs, reacts with excess oxygen? The molar mass of propane is 44.097 g/mol and the molar mass of water is 18.015 g/mol. First, determine the conversion factor that you will use to convert grams of propane to moles of propane. ﬂ_ mol C3Hg y,0970 s General Chemistry | “ Texas A&M University

Chemistry Learning Laboratory Second, determine the conversion factor that you will use to convert from moles of propane to moles of water. Y molH,0 1 molCsHg Third, determine the conversion factor that you will use to convert from moles of water to grams of water. 18.0/8g H,0 4 molH;0 Use the conversion factors above to calculate how many grams of water form when 50.0 g of propane reacts with excess oxygen. Round your answer to the correct number of significant figures. 50g GHy (1macots\ [ Ynd o\ [ 1RO58\ _(R1. 704 ] 3 '("5‘!“ 1 ol g, (""" "‘°> 3 H,_O 4. Apply what you have learned above to answer the following question. How many grams of propane react with excess oxygen, O, (MM 32.00 g/mol), to produce 38.3 grams of carbon dioxide, CO, (MM 44.01 g/mol)? 39'3, €0, [ Vel at> A ndGH) Y0077, #.0190 Imel Co, '“'Cs"n ryh&«,w Consider the following reaction when answering questions 5, 6, and 7. 2Li0OH(s) + CO5(g) - LiyC05(s) + H,0(1) 5. How many moles of lithium carbonate, Li,CO3, form when 2.5 moles of lithium hydroxide, LiOH, react with 3 moles of carbon dioxide, CO2? First, imagine that lithium hydroxide is the limiting reagent. Calculate how many moles of lithium carbonate will form if all 2.5 moles of lithium hydroxide react. 2.5 LOKml (1- '-ln@w) =| 1.25 L. c°3 mel 2 LoH mol Second, imagine that carbon dioxide is the limiting reagent. Calculate how many moles of lithium carbonate will form if all 3 moles of carbon dioxide react. 4 wel MCD. Amil COp ((— T TaChs 1, mel C o, Only one of the answers above represents the actual moles of lithium carbonate formed — the one associated with the limiting reagent. The limiting reagent is the reagent that reacts to produce the fewer number of moles of product. What is the limiting reagent? What is the excess reagent? Now that you have identified the limiting reagent, state how many moles of lithium carbonate are actually formed. limiting reagent: Ll OH excess reagent: C 02 moles Li,CO; formed: \2 g General Chemistry | “ Texas A&M University Chemistry Learning Laboratory 6. How many moles of the excess reagent in Question 5 remain after the reaction is complete? First, determine how many moles of the excess reagent react with the limiting reagent using stoichiometric conversions similar to those on the previous pages. 25L00(1 CD* 250H Subtract the moles of excess reagent that have reacted from the initial moles of excess reagent to determine the number of moles of excess reagent that remain unreacted. 2 -5 = 7. Apply what you have leamned above to answer the following question. How many moles of water, H20, form when 4 moles of lithium hydroxide, LiOH, react with 1.5 moles of carbon dioxide, CO,? What is the limiting reagent? How much of the excess reagent remains after the reaction? Nlny _. 9 15l %0\ _\ rs.nM

“uyn —7 ¢ Co T ‘}:|9 L l. - |5 K COy L{Lou("’m> =2 [02"). Consider the following reaction when answering questions 8, 9, and 10. 2NH;(g) + 3Cu0(s) = N,(g) + 3Cu(s) + 3H,0(g) 8. How many grams of nitrogen, N2 (MM 28.01g/mol), form when 22.3 grams of ammonia, NH3 (MM 17.031 g/mol), react with 35.6 grams of copper(ll) oxide, CuO (MM 79.55 g/mol)? First, imagine that ammonia is the limiting reagent. Calculate how many grams of nitrogen will form if all 22.3 grams (us,) (_I!& z N ) 2.031 Second, imagine that copper(ll) oxide is the limiting reagent Calculale how many grams of nitrogen will form if all 35.6 grarzs copper(ll) oxide reacts. 266 (Lol |...| mb \ 3 9.5404 5411 Sml GO Il General Chemistry | Texas A&M University Chemistry Learning Laboratory What is the limiting reagent? What is the excess reagent? How many grams of nitrogen, N2, are formed? limiting reagent: L (V) () excess reagent: & 953 moles N, formed: ( ) | IEI 9. How many grams of the excess reagent in Question 8 remain after the reaction is complete? First, determine how many grams of the excess reagent react with the limiting reagent using a stoichiometric conversion similar to those on the previous pages. sl (2 Subtract the grams of excess reagent that have reaaed from the initial grams of excess reagent to determine the number of grams of excess reagent that repa 22.3-Com = 10. How many grams of copper, Cu (MM 63.546 g/md vnen 10.0 grams of ammonia, NH3 (MM 17.031 g/mol), react with 80.5 grams of copper(ll) oxide, CuO (MM 79. 55 g/mol)? What is the limiting reagent? How much of the excess reagent remains after the reaction? Q= ATl C-D = 33'73 role e @;w):m;c., \05( l7033> =0 'O\ 6> 80,6‘—;#: lo.q%s‘j&p Consider the following reaction when answering questions 11, 12, and 13. 2Ca0(s) + 5C(s) - 2CaC,(g) + C0,(g) 11. When 110.0 grams of carbon, C (MM 12.011 g/mol), reacts with excess calcium oxide, CaO (MM 56.078 g/mol), 183.2 grams of calcium carbide, CaC; (MM 64.099 g/mol), forms. What is the percent yield of this reaction? The percent yield of a reaction can be calculated using the expression below.

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actual yield ——x 100 expected yield % yield = First, calculate the expected (or theoretical) yield of calcium carbide when 110.0 g C reacts with excess CaO using stoichiometric conversions similar to those on the previous pages. “03 ‘ < 2. o||55><§ml C) GZ? =234 ?Ij Now that you know the expected yield, you can calculate the percent yield. 23‘! 2 \ General Chemistry | Texas A&M University Chemistry Learning Laboratory 12. Apply what you have learned above and on the previous pages to answer the following question. Use the percent yield calculated in Question 11 to predict the actual yield when 23.5 grams of carbon react with 50.5 grams of calcium carbide. 1355 (g Ak 2859 (imC) 142 (M), p 24 Sb54 (a.mb/z -9 73/=/LIOD 13. Apply what you have learned above to answer the following question. Use the percent yield calculated in Question 11 to predict how much carbon would be needed to produce an actual yield of 85.6 grams of calcium carbide. (Hint: Find the expected yield from the percent yield and the actual yield, then work backwards to see how much C is needed). .78:%5 = I01.1g 109 Malals [ e[ 5 C 4L\ _ J 64.M1J><1 Gl fi_!g"qwfict Consider the following reaction when answering questions 14 and 15. Zn(s) + 2HCl(aq) — ZnCl,(aq) + H,(g) 14. How many grams of hydrogen, H, (MM 2.016 g/mol), are formed when 53.5 mL of a 0.150 M solution of hydrochloric acid, (HCI) reacts with excess zinc? First, determine the conversion factor that you will use to convert volume of HCI to moles of HCI. - 14D mol Hel ’L L HCl Is the volume of HCI provided in the correct units to use with this conversion factor? What do you need to do first? ND, convet mwL—> L Second, determine the conversion factor that you will use to convert from moles of HCI to moles of Hz. a4 mol H, 7. molHCL Third, determine the conversion factor that you will use to convert from moles of H, to grams of H.. 2 g Hy 4 molH; General Chemistry | G Texas A&M University

Chemistry Learning Laboratory Use the conversion factors above when determining the grams of hydrogen, Hz, produce e rpaction of 53.5 mL of 0.150 M HCI with excess zinc. 19D= 35 @Bk, () (- 1:0.00’627:\ 15. What volume of 0.250 M HCI solution, in mL, is required to react completely with 35.0 g zinc, Zn (MM 63.38 g/mol)? First, determine the conversion factor that you will use to convert moles of HCI to volume of HCI near the end. LHCL %0 mol Hcl Use the conversion factor above and stoichiometric conversions similar to those on the previous pages when determining the volume of 0.250 M HCI required to react with all of a 35.0 g sample of zinc. 359 2n (I 2n 1\ (Zrol B ) _ 1044 ol A ég?ijz \mo|2n .250=U%q‘/ 16. Apply what you have learned above to answer the following question. How many mL of a 0.150 M solution of a strong acid, H2S0s, are required to neutralize (completely react with) 300 mL of a 0.250 M solution of a strong base, NaOH? H,50,(aq) + 2NaOH(aq) - Nay$0,(aq) +2H,00) 5 oy 0l ’ 3 = 0.5 5 ¥ 0.0 7wl Mot/ Il B4 \= O.0%75 ml H,S 2 mo\ N b 4ynl = 00315 X= 5L = 0 l50= 22 General Chemistry | G Texas A&M University