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3-1 Chapter 3 Stoichiometry of Formulas and Equations Stoichiometry is derived from the Greek words στοιχε ῖ ον (stoicheion, meaning element) and μέτρον.

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Presentation on theme: "3-1 Chapter 3 Stoichiometry of Formulas and Equations Stoichiometry is derived from the Greek words στοιχε ῖ ον (stoicheion, meaning element) and μέτρον."— Presentation transcript:

1 3-1 Chapter 3 Stoichiometry of Formulas and Equations Stoichiometry is derived from the Greek words στοιχε ῖ ον (stoicheion, meaning element) and μέτρον (metron, meaning measure.)

2 3-2 화학양론 1. 몰 2. 미지 화합물의 화학식 구하기 실험식, 분자식 3. 화학반응식과 균형 맞추기 4. 반응물과 생성물의 양 한계반응물, 수득량, 수득 백분율 5. 용액에서의 화학양론 몰농도 용액 상태의 화학반응에서 화학량론

3 3-3 Mole - Mass Relationships in Chemical Systems 3.5 Fundamentals of Solution Stoichiometry 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts of Reactant and Product

4 화학전공 교수 - 학생 대화의 시간 3 월 17 일 ( 오늘 ) 오후 6:00 – 원천관 535 호실 김밥, 다과 준비 구체적인 진로지도 및 선배, 대학원생, 화학과 교수님과의 대화의 시간

5 3-5 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 79 1 mol = N A = 6.02214179(30)×10 23 Avogadro’s number (N A ) The Mole(chemical quantity)

6 3-6 겁 ( 劫 ) 과 찰라 ( 刹那 ) 찰나 [ 刹那 < 범어 k¡a¡a] ① ≪불교≫ (75 분의 1 초에 해당하는 ) 지극히 짧은 시간. 겁 ( 劫 ). ② 어떤 사물 현상이 이루어지는 바로 그때. 순간 ( 瞬間 ). 겁 [ 劫 ][< 범어 kalpa] ≪불교≫ (‘ 천지가 개벽한 때부터 다음 개벽할 때까지의 동안 ’ 이라는 뜻으로 ) ‘ 지극히 길고 오랜 시간 ’ 을 일컫는 말. 찰나. 힌두교에서는 1 겁을 43 억 2000 만 년 (1.36x10 17 초 ) 이라고 한다. =1.02×10 19

7 3-7 The Mole(chemical quantity) Avogadro’s number =6.022  10 23 만약 남북한 땅 위에 10 원짜리 동전을 쌓는다면 그 높이는 얼마나 될까 ? 10 직경 23 mm 두께 1.5 mm 녹여 쌓으면 1,700 km 사방으로 펼쳐 쌓으면 2,200 km 삼각으로 펼쳐 쌓으면 1,900 km

8 3-8 The Mole(chemical quantity) Avogadro’s number =6.022  10 23 만약 10 원 동전을 지구상 ( 땅과 바다 위 ) 에 쌓는다면 그 높이는 얼마나 될까 ? 녹여 쌓으면 730 m 사방으로 펼쳐 쌓으면 930 m 삼각으로 펼쳐 쌓으면 800 m 직경 23 mm 두께 1.5 mm

9 3-9 Molar mass is the mass of 1 mole of in grams atoms watermelons 1 mole 12 C atoms = 6.022 × 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams)

10 3-10 The Mole(chemical quantity) 10 kg 수박 1 몰 수박을 쌓아보자 !

11 3-11 6.02 × 10 24 kg ! 지구 질량 5.97 × 10 24 kg

12 3-12 One Mole of: C S Cu Fe Hg

13 3-13 Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) 6 atoms Atoms/molecule of compound 6 atoms12 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6(6.022 × 10 23 ) atoms 12(6.022 × 10 23 ) atoms 6(6.022 × 10 23 ) atoms Mass/molecule of compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu Mass/mole of compound 96.00 g 72.06 g12.10 g Oxygen (O) Carbon (C)Hydrogen (H)

14 3-14 MASS(g) of atoms MASS(g) of atoms AMOUNT(mol) of atoms AMOUNT(mol) of atoms NUMBER of atoms NUMBER of atoms mass-mole-number of atoms. M (g/mol) Avogadro’s number (atoms/mol) 79 6.02214179×10 23 atomic mass1 amu ×N A = 1 g(exact)

15 3-15 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element SOLUTION : (a) Silver (Ag) is used in jewelry and tableware. How many grams of Ag are in 0.0342 mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8 g of Fe? 0.0342 mol Ag × mol Ag 107.9 g Ag = 3.69g Ag (atomic mass(amu): Ag=107.9, Fe=55.85) 95.8g Fe× 55.85g Fe mol Fe 6.022x10 23 mol = 1.04x10 24 atoms Fe ×

16 3-16 MASS(g) of compound MASS(g) of compound AMOUNT(mol) of compound AMOUNT(mol) of compound MOLECULES (or formula units) of compound MOLECULES (or formula units) of compound AMOUNT(mol) of elements in compound AMOUNT(mol) of elements in compound Avogadro’s number (molecules/mol) M (g/mol) chemical formula mass-mole-number of compounds. molecular (or formula) mass

17 3-17 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound Ammonium carbonate is white solid that decomposes with warming. Among its many uses, it is a component of baking powder, first extinguishers, and smelling salts. How many formula unit are in 41.6 g of ammonium carbonate? SOLUTION: Ammonium carbonate (NH 4 ) 2 CO 3. M = (2 ×14.01 g/mol N) + (8 × 1.008 g/mol H) + (12.01 g/mol C) + (3 × 16.00 g/mol O) = 96.09 g/mol 41.6 g (NH 4 ) 2 CO 3 × = 2.61x10 23 (NH 4 ) 2 CO 3 mol (NH 4 ) 2 CO 3 96.09 g (NH 4 ) 2 CO 3 6.022x10 23 mol ×

18 3-18 Calculating the Mass Percents and Masses of Elements in a Sample of Compound Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating chemical potential energy. (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55g of glucose? SOLUTION: (a)M(C 6 H 12 O 6 )= 6×12.01 + 12×1.008 + 6×16.00 = 180.16 g/mol Glucose mass percent of C = 6×12.01g C/180.16 g × 100 = 39.99 mass %C mass percent of H =12×1.008g H/180.16 g ×100 = 6.714 mass %H mass percent of O =6×16.00g O/180.16 g ×100 = 53.29 mass %O (b) 16.55g Glucose ×39.99 mass %C/100 % Glucose = 6.618 g C

19 3-19 Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists. Molecular Formula = (Empirical Formula ) n

20 3-20 Determining the Empirical Formula from Masses of Elements Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? SOLUTION: 2.82 g Na/(22.99 g/mol) = 0.123 mol Na 4.35 g Cl/(35.45 g/mol) = 0.123 mol Cl 7.83 g O/(16.00 g/mol) = 0.489 mol O Na 1 Cl 1 O 3.98 → NaClO 4 NaClO 4 is sodium perchlorate. 0.123/0.123 = 1 Na 0.123/0.123 = 1 Cl 0.489/0.123 = 3.98 O

21 3-21 Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity, lactic acid ( M =90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a)Determine the empirical formula of lactic acid. (b) Determine the molecular formula. SOLUTION: Assuming there are 100. g of lactic acid, the constituents are: 40.0 g C/(12.01 g/mol) = 3.33 mol C 6.66 g H/(1.008 g/mol) = 6.66 mol H 53.3 g O/(16.00 g/mol) = 3.33 mol O CH 2 O : empirical formula C 3 H 6 O 3 is the molecular formula 3.33/3.33 = 1 C 6.66 /3.33 = 2 H 3.33 /3.33 = 1 O mass of CH 2 O molar mass of lactate 90.08 g 30.03 g =3 =

22 3-22 Combustion analyzer for CH(O) of organic compounds. concept. C n H m O x + (n + m/2 ) O 2 = n CO 2 (g) + (m/2) H 2 O (g) O2O2 H 2 O (g) absorber CO 2 (g) absorber Other gas absorber Combustion

23 3-23 Combustion analyzer for CHNS of organic compounds. Modern version. C n H m N p S r + (n + m/2 + p + r ) O 2 → n CO 2 (g) + (m/2) H 2 O (g) + p NO 2 (g) + r SO 2 (g) CHNS (O) Analyzer modern instruments are automated for analysis. 3 mg of sample is sufficient to give satisfactory CHNS analysis.

24 3-24 Determining a Molecular Formula from Combustion Analysis Vitamin C (M=176.12g/mol) is a compound of C,H, and O found in many natural sources especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO 2 after combustion=1.50 g mass of H 2 O after combustion=0.41 g What is the molecular formula of vitamin C? SOLUTION: CO 2 = 1.50 g → 1.50 g CO 2 /(44.01 g CO 2 /mol)× 12.01 g C/mol= 0.409 g C H 2 O = 0.41 g → 0.41 g H 2 O /(18.02 g H 2 O /mol)× 2.016 g H/mol= 0.046 g H O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 g O

25 3-25 C 1 H 1.33 O 1 → C 3 H 4 O 3 →88.06 g/formula mass n = 176.12 g/88.06 g = 2 (C 3 H 4 O 3 ) 2 =C 6 H 8 O 6 Determining a Molecular Formula from Combustion Analysis – cont. 0.409 g C → 0.409/12.01=0.0341 mol C 0.046 g H → 0.046/1.008=0.0456 mol H 0.545 g O → 0.545/16.00=0.0341 mol O 0.0341/0.0341 = 1 mol C 0.0456/0.0341 = 1.33 mol H 0.0341/0.0341 = 1 mol O Molecular Formula = (Empirical Formula ) n Molecular Formula mass = (Empirical Formula mass) n

26 3-26 PropertyEthanolDimethyl Ether M(g/mol)46.07 Boiling Point78.5 0 C-25 0 C Density at 20 0 C 0.789 g/mL (liquid) 0.00195 g/mL (gas) Structural formulas Space-filling models Constitutional Isomers of C 2 H 6 O

27 3-27 1.Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. 2.Set the equation for the coefficients (a, b, c & d) to make the number of atoms of each element the same on both sides of the equation. 3.Solve the equations. Usually start by setting the coefficient of the most complex compound be 1. 4.Adjust the coefficient to the most simple whole numbers. 5.Check to make sure that you have the same number of each type of atom on both sides of the equation. 6. Specify the states of matter.

28 3-28 Ethane reacts with oxygen to form carbon dioxide and water C: 2 a = c, O: 2 b = 2 c +d, H: 6 a = 2 d. 2 = c, 2 b = 2 c +d, 6 = 2 d. => d = 3 Let a = 1 6 = 2 d. => d = 3 2b = 2x2 + 3 = 7. b = 7/2 a C 2 H 6 + b O 2 → c CO 2 + d H 2 O a = 1 b = 7/2 c = 2 d = 3 a = 2 b = 7 c = 4 d = 6 X 2 2 C 2 H 6 + 7 O 2 → 4 CO 2 + 6 H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 2 C 2 H 6 (g)+ 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O(l) Balancing Chemical Equations - example

29 3-29 MASS(g) of compound A MASS(g) of compound A AMOUNT(mol) of compound A AMOUNT(mol) of compound A MOLECULES (or formula units) of compound A MOLECULES (or formula units) of compound A Avogadro’s number (molecules/mol) M (g/mol) of compound A molar ratio from balanced equation mass-mole-number relationships in a chemical reaction. MASS(g) of compound B MASS(g) of compound B AMOUNT(mol) of compound B AMOUNT(mol) of compound B MOLECULES (or formula units) of compound B MOLECULES (or formula units) of compound B Avogadro’s number (molecules/mol) M (g/mol) of compound B

30 3-30 Calculating Amounts of Reactants and Products In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide. (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units

31 3-31 SOLUTION: Calculating Amounts of Reactants and Products 2Cu 2 S( s ) + 3O 2 ( g ) → 2Cu 2 O( s ) + 2SO 2 ( g ) 3mol O 2 2mol Cu 2 S = 15.0mol O 2 = 641g SO 2 = 0.959kg O 2 kg O 2 10 3 g O 2 = 20.0mol Cu 2 O 20.0mol Cu 2 O 3mol O 2 2mol Cu 2 O 32.00g O 2 mol O 2 10.0mol Cu 2 S(a) 10.0mol Cu 2 S 2mol SO 2 2mol Cu 2 S 64.07g SO 2 mol SO 2 (b) 2.86kg Cu 2 O 10 3 g Cu 2 O kg Cu 2 O mol Cu 2 O 143.10g Cu 2 O (c) (a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? (b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted? (c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

32 3-32 Balanced Equation Viewed in Terms of Reactants → Products C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) molecules 1 molecule C 3 H 8 + 5 molecules O 2 → 3 molecules CO 2 + 4 molecules H 2 O amount (mol) 1 mol C 3 H 8 + 5 mol O 2 → 3 mol CO 2 + 4 mol H 2 O 44.09 amu C 3 H 8 + 160.00 amu O 2 → 132.03 amu CO 2 + 72.06 amu H 2 O mass (amu) mass (g) 44.09 g C 3 H 8 + 160.00 g O 2 → 132.03 g CO 2 + 72.06 g H 2 O total mass (g) 204.09 g → 204.09 g

33 3-33 Using Molecular Depictions to Solve a Limiting-Reactant Problem Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine. (a) Suppose the reaction is run with 3.00 mol of Cl 2 and 6.00 mol of F 2, name the limiting reactant, and draw the container contents after the reaction is complete. (b) What will be the mass of remaining reactants and product? F2F2 Cl 2 F2F2 ClF 3 n 0 (mol)3.006.000 n/c(mol)3/1=36/3=2 n f (mol)3-2×1=16-2×3=00+2×2=4 m f (g)1×70.9=70.9g04×92.5=370 1Cl 2 (g) + 3F 2 (g) → 2ClF 3 (g) M(g/mol)= 70.9 38.0 92.5

34 3-34 Calculating Amounts of Reactant and Product in a Limiting-Reactant Problem A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N 2 H 4 ) and dinitrogen tetraoxide(N 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00×10 2 g of N 2 H 4 and 2.00×10 2 g of N 2 O 4 are mixed? SOLUTION: 2 N 2 H 4 ( l ) + N 2 O 4 ( l ) → 3 N 2 ( g ) + 4 H 2 O( l ) 1.00×10 2 g N 2 H 4 /(32.05g/mol) = 3.12mol N 2 H 4 2.00×10 2 g N 2 O 4 /(92.02g/mol) = 2.17mol N 2 O 4 N 2 H 4 is the limiting reactant. 4.68mol N 2 ×28.02 g/mol = 131 g N 2 3.12mol N 2 H 4 /2 = 1.56 < 2.17mol N 2 O 4 /1 1.56×3 = 4.48 mol N 2 will be produced.

35 3-35 A + B C (reactants) (main product) D (side products) The effect of side reactions on yield. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 Reaction Yield

36 3-36 Calculating Percent Yield Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process? SOLUTION: SiO 2 ( s ) + 3C( s ) → SiC( s ) + 2CO( g ) 100.0 kg SiO 2 × (10 3 g /kg) ×(mol/60.09 g) = 1664 mol SiO 2 n SiO 2 = n SiC = 1664 × 100= 77.0% 51.4 kg 66.73 kg 1664 mol SiC ×(40.10 g/mol) × (kg/10 3 g) = 66.73 kg SiC

37 3-37 Calculating the Molarity of a Solution Glycine (H 2 NCH 2 COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL? SOLUTION: 0.715 mol glycine 495 mL soln 1000mL 1 L = 1.44 M glycine Molarity ( M, mol/L) = n solute /V solution

38 3-38 mass-mole-number-volume relationships in solution. MASS (g) of compound in solution AMOUNT (mol) of compound in solution MOLECULES (or formula units) of compound in solution VOLUME (L) of solution M (g/mol) Molarity (mol/L) Avogadro’s number (molecules/mol)

39 3-39 Calculating Mass of Solute in a Given Volume of Solution A “buffered” solution maintains acidity as a reaction occurs. In living cells phosphate ions play a key buffering role, so biochemistry often study reactions in such solutions. How many grams of solute are in 1.75 L of 0.460 M sodium monohydrogen phosphate? SOLUTION: 1.75 L× 0.460 moles 1 L 0.805 mol Na 2 HPO 4 × 141.96 g Na 2 HPO 4 mol Na 2 HPO 4 = 0.805 mol Na 2 HPO 4 = 114 g Na 2 HPO 4

40 3-40 Converting a concentrated solution to a dilute solution. The total number of solutes in the solution remains the same after dilution, But the number of solutes per unit volume changes...

41 3-41 Preparing a Dilute Solution from a Concentrated Solution “Isotonic saline” is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? the number of moles of solute does not change during the dilution but the volume does M dil × V dil = n solute = M conc × V conc SOLUTION: n solute required: 0.80 L soln ×0.15 M NaCl = 0.12 mol NaCl volume of conc. solution = 0.12 mol NaCl/6M = 0.020 L soln

42 3-42 Calculating Amounts of Reactants and Products for a Reaction in Solution Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10 g of magnesium hydroxide? SOLUTION: Mg(OH) 2 (s) + 2HCl(aq) → MgCl 2 (aq) + 2H 2 O(l) n Mg(OH) 2 /1 = n HCl/2 0.10g / (58.33g/mol) = 1.7 × 10 -3 mol Mg(OH) 2 = n HCl/2 n HCl = 3.4 × 10 -3 mol HCl = 0.10 M × V HCl V HCl = 3.4 × 10 -2 L HCl

43 3-43 Solving Limiting-Reactant Problems for Reactions in Solution Mercury and its compounds have many uses, from fillings for teeth (as an alloy with silver, copper, and tin) to the industrial production of chlorine. Because of their toxicity, however, soluble mercury compounds, such mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050L of 0.010 M mercury(II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams of mercury(II) sulfide form? SOLUTION: Hg(NO 3 ) 2 ( aq ) + Na 2 S( aq ) → HgS( s ) + 2NaNO 3 ( aq ) 0.050L×0.010M = 5.0×10 -4 mol HgS 0.020L×0.10M = 2.0x10 -3 mol Na 2 S Since (5.0×10 -4 mol Hg(NO 3 ) 2 /1) < (2.0x10 -3 mol Na 2 S /1), Hg(NO 3 ) 2 is the limiting reagent. (5.0×10 -4 mol ) ×1 HgS will form. 5.0×10 -4 mol × 232.7g/mol = 0.12 g HgS

44 3-44 aA + bB → cC + dD Limiting Reagent and theoretical Yield Reactants Products

45 3-45 1. 다음 중 원소기호와 그 이름이 바르게 짝지어지지 않은 것을 고르고 바른 이름을 쓰시오. (a) Zn - Iron (b) K - Potassium (c) Sn - Antimony (d) Au - Gold (e) Ce - Cesium (f) Hg - Hydrogen (g) Fe - Steel (h) Ar -Arsenic

46 3-46 2. 다음의 순물질 중에서 원소가 아닌 것을 고르시오. 원소가 아닌 것의 이유를 쓰시오. (a) 수정 (b) 다이아몬드 (c) 흑연 (d) 풀러렌 (C 60 ) (e) 니켈 (f) 오존 (g) 사파이어 (i) 황동 (j) 청동 (h) 테플론 (j) 순금 (k) 브로민 (l) 주석 (m) 백금 (n) 지르코늄

47 3-47 메테인 (CH 4 ) 18.5 g 이 염소 43.0 g 과 반응하여 80% 수득률로 염화 메틸 (CH 3 Cl) 이 생성되었다. 생성된 반응물은 몇 g 인가 ? 이 때 염화 수소도 같이 생성되었다. 원자질량 (amu): C: 12.01, H:1.008, Cl: 35.45

48 3-48 뷰테인 ( 부탄 ) 기체는 압축시켜서 일회용 라이터나 휴대용 가스버너의 액체 연료로 사용한다. 라이터에 뷰테인 6.50 mL 가 들어있다 ( d = 0.579 g/mL ). (a) 뷰테인을 전부 연소시키기 위해 필요한 최소량의 산소는 몇 g 인가 ? (b) 뷰테인이 전부 연소하면 CO 2 는 몇 mol 이 생성되겠는가 ? (c) 뷰테인이 전부 연소하면, 전체 기체 분자는 얼마나 되겠는가 ? 원자질량 (amu): C: 12.01, H:1.008, O: 16.00

49 3-49 진한 황산 (18.3 M ) 은 밀도가 1.84 g/mL 이다. (a) 1 L 용액 안에 황산 몇 mol 이 들어있는가 ? (b) 용액에서 H 2 SO 4 의 질량 백분율은 얼마인가 ? 원자질량 (amu): C: 12.01, H:1.008, O:16.00, S: 32.07


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