1. Balance the following equations. Al 2 (SO 4 ) 3 (aq) + 3 Ba(NO 3 ) 2 (aq) ----> 2 Al(NO 3 ) 3 (aq) + 3 BaSO 4 (s)

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1 Stoichiometry: Calculations with Chemical Formulas and Equations (Chapter ) Past Quiz and Test Questions Answer Key My answers are in bold faced italics and underlined to this point I have only included answers up to question 9.. Balance the following equations Al (SO ) (aq) + Ba(NO ) (aq) ----> Al(NO ) (aq) + BaSO (s) C H 8 O (R) + O (g) > CO (g) + H O (g) P O 5 (g) + H O (R) ----> H PO (aq) B 0 H 8 + O -----> 5B O + 9 H O N (g) + H (g) -----> NH (g). a. How many grams of Cl are in.5-g of C Cl 6?.5gC Cl6 gcl? gcl 0. 0gCl 7gC Cl b. How many atoms of O are in 5.0-g of C H O? 5.0gCH O 6.00 atomso? atomso 8.70 gc H O 6 atomso c. How many moles of CaBr are required to obtain.5 0 atoms of Br?.50 atomsbr molcabr 6.00 atomsbr? molcabr.70 molcabr d. How many grams of C are required to make.5-mol of CH?.5molCH gc? gc 6. gc molch e. How many moles of NH are contained in 50.0-g of NH? 50.0gNH molnh? molnh. 7molNH 7gNH

2 . Consider the following chemical equation. Given: 0.0 g K CO 0.0 g AlBr K CO (aq) + AlBr (aq) > 6 KBr (aq) + Al (CO ) (s) Stoichiometry: mol K CO mol AlBr 6 mol KBr mol Al (CO ) 8 g K CO 67 g AlBr 6 9 g KBr g Al (CO ) 0.0-g of K CO are reacted with 0.0-g of AlBr. a. Balance the equation. b. Which of the reactants is the limiting reactant? Compare 0.0gK CO 8gK CO Since the stoichiometric ratio involving K CO is less than that involving AlBr, K CO is the limiting reactant. c. How many grams of Al (CO ) can be formed? 0.0gKCO gal( CO )? gal ( CO ).gal( CO ) 8gKCO Note this calculation is based on the limiting reactant, not the ecess. d. Suppose the reaction is carried out and 7.5-g of Al (CO ) are formed. What is the percent yield? 7.5gAl (CO) Produced % yield 00% 66.% yield.gal (CO ) Calculated gAlBr 67gAlBr e. How much of the ecess reactant remains after the reaction if all of the limiting reactant is consumed? First, figure out how much of the ecess reactant, AlBr, was used. 0.0gKCO 67gAlBr? galbrused 5. 8gAlBr used 8gK CO Ecess AlBr 0.0 g 5.8 g. g of ecess reactant left over after the reaction

3 . Determine the empirical and molecular formulas for epinephrine, a hormone secreted into the blood stream during times of danger or stress. Its elemental composition is 59.0% C, 7.% H, 6.% O, and 7.7% N. Its molecular weight is 8. C : Divide by smallest(0.55) : H : O : N : Since these are all close to an integer, round to the nearest integer for a ratio of: 9::: or C 9 H O N for the empirical formula. Since the formula weight of the empirical formula is 8 and matches the molecular weight, the molecular formula is also C 9 H O N. 5. Balance each of the following equations. C H O (s) + O -----> CO (g) + H O (R) Mg (s) + HCl (aq) -----> MgCl (aq) + H (g) Na CO (aq) + HNO (aq) ----> NaNO (aq) + H CO (aq) SO (g) + H O (R) -----> H SO (aq) KHCO (s) ----> K CO (s) + CO (g) + H O (g)

4 6. a. How many moles of BaBr are contained in 50.0-g of BaBr? 50gBaBr molbabr? molbabr 0. 69molBaBr 7gBaBr b. How many atoms of Cl are in.6-g of CHCl?.6gCHCl 6.00 atomscl? atomscl gCHCl atomscl c. How many grams of Sr are contained in sufficient SrCl to obtain 7. 0 atoms of Cl? 7.0 atomscl 87.6gSr gsr 5. gsr 6.00 atomscl? d. How many grams of N are contained in 50 molecules of N H?? gn 50moleculesNH 8gN moleculesn H 0 gn e. How many molecules of H SO are contained in 75.0-g of H SO? 75.0gH SO 6.00 moleculesh? moleculesh SO.60 moleculeshso 98gH SO SO

5 7. a. How many grams of N are there in 5.0-g of NH NO? 5.0gNH NO 8gN? gn 5. 5gN 80gNH NO b. How many molecules of water are in one gallon - (or.78 L). Assume the density of water is g/ml..78l 000mL g 6.00 moleculesho 6? moleculesho.660 moleculesho L ml 8gH O c. How many moles of C H 6 are in 5.0-g of C H 6? 5.0gC H molc H? molc H molc H gCH 6 6 d. How many atoms of oygen are contained in 5.0-g of KMnO? 5.0gKMnO 6.00 atomso? atomso gKMnO atomso e. How many grams of BaCl are required to obtain 50 atoms of Cl? 50atomsCl 08gramsBaCl 6.00 atomscl 0? gramsbacl 7.80 gramsbacl

6 8. Hydrazine, N H, is used etensively as a rocket fuel. It can be prepared from the reaction: Given: 5.0-g NH 7.5-g OCl - NH (g) + OCl - (aq) ----> N H (aq) + Cl - (aq) + H O (R) Stoichiometry: mol NH mol OCl - mol N H mol Cl - mol H O 7 g NH 5.5 g OCl - g N H 5.5 g Cl - 8 g H O Suppose the reaction is carried out with 5.0-g of NH and 7.5-g of OCl -. a. Which is the limiting reactant? 5.0gNH 7.5gOCl Compare : gNH 5.5gOCl Since the stoichiometric ratio for OCl - is less than that for NH, OCl - is the limiting reactant. b. How many grams hydrazine could be formed in the reaction above? Based on limiting reactant: 7.5gOCl gnh? gn H 0. 9gNH 5.5gOCl c. In the above reaction, how many grams of the ecess reactant would be epected to remain after the reaction is complete. First, find out how many grams of the ecess reactant, NH, reacted. 7.5gOCl 7gNH? gnh reacted. 6gNH 5.5gOCl The remaining amount is the starting amount minus the reacted amount: 5.0-g.6g.g NH remain

7 9. The reason hydrofluoric acid, HF(aq), cannot be stored in glass bottles and can be used in etching glass is its reaction with silica, SiO. 5.0-g SiO 5.0-g HF SiO (s) + 6 HF (aq) W H SiF 6 (aq) + H O (l) mol SiO 6 mol HF mol H SiF mol H O 60 g SiO 6 0 g HF g H SiF 8 g H O Suppose 5.0-g of SiO are reacted with 5.0-g of HF. a. Which is the limiting reactant? Using conversion factor approach: 5.0gSiO 60gHF? g HF required to use up SiO 0. 0g HF required 60gSiO Since we only have 5.0-g of HF, we will run out and HF will be the limiting reactant. Alternatively, look at the two ratios formed from the equation above: 5.0gSiO 5.0gHF gSiO 60gHF Since the HF ratio is smaller, it is limiting using this approach, too, which is fortunate. b. How many grams of H SiF 6 could be produced in the reaction? 5.0gHF g H SiF6? g H SiF6 formed 0. 0g H SiF6 formed 60gHF Notice this calculation is based on the limiting reactant, HF. It is also an absolute coincidence that the grams of HF required and the grams of H SiF 6 are the same c. If the eperiment actually produces.5-g of H SiF 6 what is the percent yield? actual yield.5g % yield 00% 00% 75.0% theoretical yield 0.0g yield d. How many grams of the ecess reactant remain at the end of the reaction? 5.0gHF 60gSiO? SiO used. 60gHF g 5g SiO used #g SiO remaining at end 5.0 g started with.5 g used.5 g SiO

8 0. The following reaction occurs during the production of ammonia g N 0.0 g H N (g) + H (g) W NH (g) mol N mol H mol NH 8 g N g H 7 g NH a. Balance the equation. b g of N are mied with 0.0-g of H. Which reactant is the limiting reactant? First, find out how much H is required to react with the N, or vice versa if you wish. 55.0gN gh? g H required to react with 55.0g N. 8g H required 8gN Since we have 0.0 g of H and only need.8 g, we have ecess hydrogen and the nitrogen limits the reaction. c. How many grams of ammonia could be produced from the reaction in part b? Using the limiting reactant, N : 55.0gN 7gNH? g NH produced 66. 8g NH 8gN produced d. How many grams of the ecess reactant would remain at the end of the reaction? From part b, we know that.8 g of H were used in the reaction. To find the ecess, simply subtract.8g from the starting amount. #g H left over 0.0 g.8 g. g H

9 . Balance each of the following equations by placing coefficients in the appropriate places. C 6 H 6 (R) + 5 O W CO (g) + 6 H O (R) FeO (s) + O (g) W Fe O (s) Au S (s) + H (g) W Au (s) + H S (g) KClO (s) W KCl (s) + O (g) Eu (s) + 6 HF (g) W EuF (s) + H (g). Find the number of moles and molecules in 7.50-g of CO. Be sure to show your work. 7.50gCO molco? moles molCO gco 7.50gCO 6.00 moleculesco? molecules.950 molecules CO gco. Find the number of grams and moles in.5 0 formula units of KBr.. The reusable booster rockets of the space shuttle program use a miture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is: 000 g NH ClO Al (s) + NH ClO (s) W Al O (s) + AlCl (s) + NO (g) + 6 H O (g) mol Al mol NH ClO mol Al O mol AlCl mol NO 6 mol H O 7 g Al 7.5 g NH ClO 0 g Al O.5 g AlCl 0 g NO 6 8 g H O a. What mass of aluminum is required to react with every 000-g of ammonium perchlorate burned? 000gNH ClO 7gAl? gal 9. 8gAl 7.5gNHClO b. What mass of NO is produced when the 000-g of ammonium perchlorate is burned? 000gNH ClO 0gNO? gno 55. gnoproduced 7.5gNH ClO

10 5. a. How many N atoms are contained in 75.0-g of N H known as the rocket fuel hydrazine? 75.0gNH 6.00 Natoms? Natoms.80 gn H Natoms b. How many hydrogen atoms are contained in 00 molecules of H O? hydrogen atoms in each molecule, 00 molecules 00 hydrogen atoms 00 hydrogen atoms c. How many moles of CHF Cl are contained in 50.0-g of CHF Cl? 50.0gCHFCl molchfcl? molchfcl. 7molCHFCl 86.5gCHF Cl

11 6. Ammonia can be converted to nitric oide, NO, by the reaction below: 5.0 g NH 5.0 g O NH (g) + 5 O (g) ----> NO (g) + 6 H O (g) mol NH 5 mol O mol NO 6 mol H O 7 g NH 5 g O 0 g NO 6 8 g H O a. 5.0-g of NH are reacted with 5.0-g of O. Which is the limiting reactant? Figure out how many grams of O are required to use up all of the NH. 5.0g NH 5g O? g O 5. g O 7g NH required Since we need 5. g of O to react with all of the NH and only have 5.0g of O, we will run out of O. It is limiting. b. How many grams of NO could be formed in the reaction of part a? 5.0g O 0gNO? g NO 8. 8 g 5g O NO c. How many grams of the ecess reactant would remain after the reaction of part a? First, find out how many grams of the ecess reactant, NH, are used. 5.0 g O 7g NH? g NH used 0. 6g NH used 5g O Amount of NH remaining start amount of NH amount of NH used 5.0 g 0.6 g. g NH d. If 5.0-g of NO is formed in the reaction of part a, what is the percent yield? actual yield 5.0gNO % yield 00% 00% 79.8% theoretical yield 8.8gNO

12 7. Balance each of the following equations and classify it as combustion, decomposition, combination, methathesis, and/or oidation-reduction. Note that some equations may fall into more than one category. State all categories that apply in each case. Equation Classification C(gr) + O (g) > CO (g) combination,oid-red NH (g) + HCl (g) > NH Cl (s) Cr(s) + NiSO (aq) ----> Ni (s) + CrSO (aq) HC H O (aq) + NaOH(aq) >HOH (R) + NaC H O (aq) combination oidation-reduction metathesis H O (R) > H (g) + O (g) decomposition, oidation-reduction 8. Answer each of the following questions. Show your work to receive full credit. a. How many moles are contained in.5-g of CH Br?.5g CHBr mol CHBr? mol CHBr 0. 5 mol CH Br 95g CH Br b. How many oygen atoms are contained in 5 molecules of C H 8 O? 5 molecules C H8O O atoms? mol O atoms 0 O atoms molecule C H O or just think about it O atoms per molecule and 5 molecules 0 O atoms c. How many grams are contained in 0.5-mol of Ba(NO )? 0.5 moles Ba( NO ) 6 g Ba( NO )? g Ba( NO NO ).6 g Ba( ) mole Ba( NO ) 8 d. What is the molar mass of MgSO C7H O? molar mass g/mol e. How many grams of carbon are contained in 5.0-g of CH Cl? 5.0 g CH Cl g C? g C. 5g C 85gCH Cl

13 9. Balance each of the following chemical equations by placing the appropriate coefficients on the underlines. TiCl + H O 6 TiO + HCl P + 0 Cl 6 PCl 5 Al O + H SeO 6 Al (SeO ) + H O C H O 6 CO + 6 H O Ca (PO ) + H SO 6 Ca(H PO ) + CaSO 0. Nitromethane, CH NO, reacts with oygen to form carbon dioide, water, and nitrogen dioide. a. Write the balanced equation for this reaction. CH NO + 7 O CO + 6 H O + NO b. How many moles of oygen would be required to react with 8-mol of nitromethane? It takes 7 moles of O for every mol of nitromethane, so 8 mol of nitromethane would require 8 mol 7/ mol O.. The $-blocker drug, timolol, is epected to reduce the need for heart bypass surgery. Its composition by mass is 7. %C, 6.55 %H,.0 %N, 5.9 %O, and 7. % S. Its molar mass is -g/mol. a. What is the empirical formula for timolol? Divide by smallest: C: 7./.9.9/ C 7 H 8 N O 7 S H: 6.55/ /0. 8. N:.0/ /0..00 Mass of this formula O: 5.9/ / S: 7./ 0. 0./0. b. What is the molecular formula for timolol? Since the molar mass mass of the empirical formula, the empirical formula is the molecular formula: C 7 H 8 N O 7 S

14 . Find the percent mass of each element in C 9 H 9 N. 9 % C 00% 8.% C % H 00% 6.87% H % N 00% 0.7% N a. How many grams of NH are contained in mol of NH?.50 mol NH 7g NH? g NH 0. 07g NH mol NH b. How many atoms of C are contained in 0.9-g of C H O? 0.9g C H O 6.00 atoms C? atoms C.08 g C H O 0 c. How many moles of PCl are contained in 75.0-g of PCl? 75.0g PCl mol PCl? mol PCl mol PCl 7.5 g PCl d. How many atoms of oygen are contained in 5 molecules of CO? 5 molecules CO atoms O? atoms O 70 atoms O molecule CO atoms C e. How many molecules of H O are contained in 5.0-g of water? 5.0 g H O 6.00 molecules H O? molecules H O.70 molecules H O 8 g H O. a. How many grams of BaI are contained in.5-mol of BaI?.5 mol BaI 9 g BaI? g BaI 89 g BaI mol BaI b. How many oygen atoms are contained in 0.0-g of water? 0.0 g H O 6.00 O atoms? O atoms.50 8 g H O O atoms c. How many moles of NH NO are contained in 5.0-g of NH NO? 5.0 g NH NO mol NH NO? mol NH NO 0. mol NH NO 80 g NH NO