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UNASANA COLLEGE MOLE AND CHEMICAL CALCULATIONS WORKSHEET FOR 2ND GRADES MOLE NUMBER n=m/mw or Aw n=V/22,4 ( at STP, for gases) n=NA or Nm / 6,02.1023 Ex: Calculate the molecular weights of a) sodium hydroxide b) calcium carbonate c) nitric acid d) water e) sodium carbonate f) potassium permanganate g) silver nitrate h) acetic acid solution: a)Mw ( NaOH)= 23+16+1=40g/mol b) Mw (CaCO3)= 40+12+3.16=100g/mol c) Mw (HNO3) =1+14+3.16=63 g/mol d) Mw (H2O) =2.1+16=18 g/mol e) Mw ( Na2CO3)=23.2+12+3.16=106 g/mol f) Mw (KMnO4) =39+55+4.16=158 g/mol g) Mw (AgNO3)=108+14+3.16=170 g/mol h) Mw (CH3COOH)=2.12+2.16+4=60 g/mol Ex : Find the mole number of followings… a) 112 g. of Iron b) 30 g. of acetic acid c) 33.6 L of carbondioxide gas at STP d) 12,04.1022 water molecules Solution: a) b) c) d) n=112/56 = 2 moles n=30/60 = 0.5 mole n=33.6/22.4 = 1.5 mole n= 12,04.1022/6,02.1023 = 0.2 mole Ex: What is the mass of 11.2 L of sulphur dioxide gas at STP ? n= 11.2/22.4 = 0.5 mole 0.5= m/64 → m= 32 g. Ex : How many molecules are there in 72 g of water ? n= 72/18 = 4 moles 4=Nm/6,02.1023→ Nm = 24,08.1023 Ex : 0.05 mol X2O3 weights 5.1 g calculate the atomic weights of X ? 0.05 = 5.1/ Mw → Mw = 102 → 2.X+3.16 → X= 27 g/mol (Al ) Ex: A 0.4 mol mixture of nitrogen monoxide and nitrogendioxide gases weights 16.8 g what is the mole percentage of nitrogen monoxide in the mixture NO x mol NO2 UNASANA COLLEGE y mol x+y =0.4 30x + 46y = 16.8 16y=4.8 → y=0.3 ,x=0.1 Page 1 Ex: What is the mass percentage of carbon in acetic acid ? M(CH3COOH)=60 g/mol mC=24 %(C) = 24/60.100=%4 Ex: If 40% of XO3 is X by weight , what is the molecular weight of the compound ? 60% 48 100% X X=80 g/mol Reaction Review: t Decomposition: CaCO3 → CaO + CO2 t KClO3 → KCl + 3/2 O2 H2O HgO Elect. → H 2 + ½ O2 t → Hg + ½ O2 Synthesis : 2 Al + 3 Cl 2 → 2AlCl3 4 Na + O2 → 2Na2O CaO + H2O → Ca(OH)2 SO3 + H2O → H2SO4 Single Displacement: 2KCl + F2 → 2KF + Cl2 KF + Cl2 → no CuCl2 + Zn → ZnCl2 + Cu Double Displacement: AgNO3 + HCl → AgCl ↓ + HNO3 BaCl2 + H2SO4→ BaSO4 ↓ + 2 HCl AgNO3 + KI → AgI ↓ + KNO3 STOICHIOMETRIC CALCULATIONS Mole calculations related to coefficients of a chemical equation is called STOICHIOMETRIC CALCULATION. xA + yB x molecules → zC y molecules z molecules x mol of A react with y mol of B produce z mol of C example: 2H2 + O2 2H20 UNASANA COLLEGE Page 2 Ex: 2C + O2 → 2CO 2 moles of carbon react with 1 mole of oxygen to form 2 moles of carbon monoxide. Ex : When 2 moles of sodium are put into water , what volume of hydrogen will form at STP ? Na 1 mol 2 mol + H2O → NaOH + ½ H2↑ 0.5 mol 1 mol =22.4 L Ex: How many liters of carbon dioxide gas can be obtain from the decomposition of 50 g calciumcarbonate at STP ? CaCO3 CaO +CO2 t CaCO3 → CaO + CO2↑ n =50/100 n = 0.5 = V/22.4 → V= 11.2 L = 0.5 Ex: Calculate the volume of air needed for the combustion of 23 g ethanol C2H5OH ? (air contains nearly 20% oxygen by volume ) (C:12 , O:16, H:1 ) C2H5OH CO2 + H2O (unbalanced equation) C2H5OH + 3O2 → 2CO2 + 3H2O n = 23/46 1.5 mol = 0.5 1.5= Vo2/ 22.4 → Vo2 = 33.6 L Vair = 5 . 33,6 = 168.0 Ex : 54 g sample of Aluminum is reacted with hydrochloric acid ; a) How many molecules of Aluminum chloride can form ? b) How many liters of gas can form at STP ? Al + HCl AlCl3 + H2 (unbalanced equation) 2Al + 6 HCl → 2AlCl3 + 3H2 ↑ 2 moles UNASANA COLLEGE 2 moles nAl=54/27=2 mol 3 moles Page 3 1 mol 6,02.1023 2 mol X x x= 12,04.1023 molecules of AlCl3 1 mol 22.4 L 3 mol x= 67.2 L H2 Note : We can use volume coefficient relationship if substances are gases. N2(g) + 3H2(g) → 2 NH3 (g) 1L 3L 2L 1 mol 3 mol 2 mol Ex:Calculate the volume of hydrogen and nitrogen to produce 60L of ammonia ? N2 + 3H2 → 2 NH3 30 L 90 L 60 L Limiting Reagent In a chemical reaction, the limiting reagent is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent. 4 chairs 5 students ( each can take 1 students) 4 chairs 5 students limited excess excess part does not react LIMITING & EXCESS PROBLEMS Ex: When 2 moles of hydrogen and 2 moles of oxygen are reacted to give water, which one becomes excess reagent? How many grams of water can be obtained at most? 2H2 + initial: 2 moles reacted:2moles excess: limit O2 → 2 moles 1mol 1 mol excess 2 H2O ----2 moles formed mH2O=2*18=36 g Ex: 18 g of carbon and 8 g of oxygen are reacted to form carbondioxide wwhich element and how many grams remain behind UNASANA COLLEGE Page 4 C + O → CO2 nc= 18/12=1.5 mol nO2=8/32=0.25 mol 1.5 mol 0.25 mol 1.25 mol of carbon is excess 1.25= m/12 →m=15g Ex: If 20 L of nitrogen and 15 L of hydrogen gases are reacted, which gas will be excess? And how many liters? N2 + 3 H2 → 2NH3 1L N2 3L H2 5L 15L XL 15 L H2 15L nitrogen excess X= 5L Ex: Equal weights of sodium and chlorine are reacted to give 0.1 mol of sodium chloride. Which element remain behind ? and how many grams ? 2Na + Cl2 → /23 →mNa = 2.3g 0.1mol 0.05mol →mCl2=3.55g initial excess 3.55g mexcess Na 2 NaCl 0.1mol 0.1= mNa 0.05=mCl2/71 3.55g = 3.55- 2.3 = 1.25g CONSECUTIVE REACTIONS Some substances can be obtained in a few reactions. Ex: can be steps. Ex: the S + O2 → SO2 2SO2 + O2 → 2SO3 As you see sulphuric acid SO3 + H2O → H2SO4 obtained from sulphur in 3 How many liters of oxygen at STP will be needed to burn methane obtained from 14,4 g of aluminum carbide? Al4C3 + 12H2O → 4 Al(OH)3 + 3 CH4 0.1 0.3 mol CH4 + 0.3 mol 2O2 → CO2 + 2H2O 0.6 mol 0.6=V/22.4 → V=13.44 L Ex: Nitric acid can be obtained in three steps from ammonia nitric 4NH3 + 5O2 → 4NO + 6 H2O 0.5 mol 0.5mol 11.2 L of 2NO + O2 → 2NO2 0.5mol 0.5 mol 4NO2 + O2 → 4HNO3 mol UNASANA COLLEGE How many grams of acid can be from ammonia at STP nNH3=11.2/22.4=0.5 Page 5 0.5mol mHNO3=31.5 g 0.5 mol 0.5=m/63 → Ex : Synthetic alcohol is produced according to the reactions ; CaC2 + 5H2O → Ca(OH)2 + C2H2 2 moles 2 moles C2H2 + H2O → CH3CHO 2moles 2 moles CHJ3CHO + H2 → CH3CH2OH 2moles 2 moles Ex: How many grams of ethanol that is 92% pure , can be obtained from 160 g of calcium carbide that is 80% pure ? mpureCaC2 = 160.80/100=128g nCaC2 = 128/64=2 moles m pure alcohol = 2 . 46 =92 g 92% alcohol 100% alcohol X= 100 g 92 g X EMPIRICAL FORMULA DETERMINATION Empirical = simplest C6H12O6 ----- molecular formula CH2O --------- empirical formula By empirical formula we know the ratios of atoms only Ex: A 50 g of compound contains 20 g of calcium , 6 g of carbon and 24 g of oxygen. Find the empirical formula of this compound. m n CaCO3 n Ca 20 20/40 0.5/0.5 1 C 6 6/12 O 24 24/16 0.5/0.5 1 1.5/0.5 3 Ex: An oxide of sulphur contains 50% by mass. Find its empirical formula. S m n SO2 n 50 50/32 1.56/1.56 UNASANA COLLEGE O 50 50/16 3.12/1.56 Page 6 1 2 Ex: A compound contains 43.40 % sodium, 11.32 % carbon, 45.28 % oxygen by mass. Find the formula of compound and write its reaction with hydrochloric acid. Na C m 43.4 n 43.4/23 Na2CO3 n 1.88/0.94 2 O 11.32 11.32/12 45.28 45.28/16 0.94/0.94 1 2.82/0.94 3 Na2CO3 + 2HCl → 2NaCl + CO2 + H2O * 1 molecule water contains 2 atoms of hydrogen and 1 atom of oxygen * 1 mole of water contains 2 moles of hydrogen and 1 mole of oxygen Ex:16 g of Ca( XO3)2 contain 0.2 mole of X. Whaat is the atomic weight of X? nX = 0.2 so nCa(XO3)2 = 0.1 = 16.4/Mw Mw= 164 = 40+2x +6.16 ---- x=14 g/mole N Ex: A 2.2 g compound containing carbon and hydrogen is burned and 3.36 L of carbon dioxide at STP are produced . What is the empirical formula of the compound? CxHy + O2 → CO2 + =3.36/22.4= 0.15 mole 2.2 g 6.6 g 0.15*4.4 = 6.6 g 44 g CO2 12g C 6.6 g CO2 0.4 mC = 1.8 g 0.4/1 xg H H2O nCO2 mCO 2 = C m mH = 0.4 g 1.8 n n 1.8/12 0.15/0.15 0.4/0.15 1 2.6 C3H8 Ex : When a 2.3 g compound containing carbon , hydrogen and oxygen is burned 4.4 g of carbon dioxide and 2.7 g of water are produced . Find the formula of the compound? CxHyOz + O2 → CO2 + H2O 44 g CO2 4.4 g mC =1.2 g 1.2-0.3=0.8g C 12 g C x H UNASANA COLLEGE 18 g H2O 2.7 g m H = 0.3 g 2gH x mO=2.3- O Page 7 1.2 1.2/12 0.1/0.05 2 0.3 0.3/1 0.3/0.05 6 0.8 0.8/16 0.05/0.05 1 C2H6O Ex : The combustion of 0.25 mole of compound containing carbon hydrogen and oxygen requires 1.25 mole oxygen and produces 1mole of carbondioxide and 1 mole of water.What is the molecular formula compound ? CxHyOz + 5O2 → 4CO2 + 4 H2O 0.25mol 1.25mol x=4 y=8 z=2 1mol 1 mol → multiply by 4 C4H8O2 MORE QUESTIONS 1. The fermentation of glucose, C 6H12O6, produces ethyl alcohol, C2H5OH, and carbon dioxide. C6H12O6 (aq) 2C2H5OH(aq) + 2CO2(g) How many grams of ethanol can be produced from 10.0 g of glucose? 2. Silicon carbide, SiC, is commonly known as carborundum. This hard substance, which is used commercially as an abrasive, is made by heating SiO2 and C to high temperatures. SiO2(s) + 3C (s) SiC (s) + 2CO(g) How many grams of SiC form when 3.00 g of SiO2 and 4.50 g of C are allowed to react? 3. What mass of silver chloride can be made from the reaction of 4.22 g of silver nitrate with 7.73 g of aluminum chloride? (Be sure to balance the reaction.) AgNO3 + AlCl3 Al(NO3)3 + AgCl 4. How many grams of iron oxide, Fe2O3 , can be produced from 2.50 g of oxygen O2 , reacting with solid iron Fe ? UNASANA COLLEGE Page 8 5. How many moles of H2O are produced when 2.5 mol of O2 react according to the following equation? C3H8 + 5O2 3CO2 + 4H2O 6. Octane burns according to the following equation. C8H18 + O2 CO2 + H2O How many grams of CO2 are produced when 5.00 g of C8H18 are burned? 7. Nitrogen gas reacts with hydrogen gas to produce ammonia gas. N2(g) + 3 H2(g) 2 NH3(g) What volume of H2 is required to react with 3.00 L of N2, and what volume of NH3 is produced at 200°C? 8. A byproduct of the reaction that inflates automotive airbags is very reactive sodium, which can ignite in air. Sodium produced during the inflation process reacts with another compound added to the airbag contents, KNO 3, via the reaction 10Na + 2KNO3 K2O + 5Na2O + N2 How many grams of KNO3 are required to remove 5.00 g of Na? 9. The alcohol in "gasohol" burns according to the following equation. C2H5OH + 3O2 2CO2 + 3H2O How many moles of CO2 are produced when 3.00 mol of C2H5OH is burned in this way? 10. Automotive airbags inflate when sodium azide, NaN 3, rapidly decomposes to its component elements via this reaction. NaN3 Na + N2 How many grams of sodium azide are required to form 7.00 g of nitrogen gas? UNASANA COLLEGE Page 9 11. CO2 exhaled by astronauts is removed from the spaceship atmosphere by reaction with KOH. CO2 + 2KOH K2CO3 + H2O How many kg of CO2 can be removed with 1.00 kg of KOH? 12. In the reaction 3NO2 + H2O 2HNO3 + NO, how many grams of HNO3 can form when 1.00 g of NO2 and 2.25 g of H2O are allowed to react? 13. In making H2O from hydrogen and oxygen, if we start with 4.6 mol of hydrogen and 3.1 mol of oxygen, how many moles of water can be produced and what remains unreacted? 14. How many grams of H2O are formed from the complete conversion of 32.00 g O2 in the presence of H2, according to 2H2 + O2 2H2O? 15. A reaction that produces crude iron from iron ore is shown below. Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) How many moles of iron are produced from the reaction of 10 mol Fe2O3 and 25 mol of CO? 16. In the synthesis of ammonia gas from nitrogen and hydrogen, N2 (g) + 3H2 (g) 2NH3 (g), what is the maximum volume of NH3 that can be formed from 10.00 L of H2? 17. If 3.00 mol of gaseous SO2 react with oxygen to produce sulfur trioxide, how many moles of oxygen are needed? Percent Yield: Write the balanced chemical equation, and SHOW ALL WORK to determine the percent yield. 18. In the reaction 2 NH3 (g) → 3 H2 (g) + N2 (g) , if 12.0 g of ammonia produced 1.87 g of hydrogen, what was the percent yield? 19a) Excess hydrochloric acid was reacted with 8.57 g of calcium carbonate and produced 3.11 g of CO2. What was the percent yield? b) In another experiment, using 9.21 g of calcium carbonate, 1.90 L of CO2 was collected. What is the percent UNASANA COLLEGE Page 10 yield? 20. What is the percent yield of precipitate if a solution containing 33.4 g of sodium phosphate produced 19.6 g of precipitate when reacted with excess aluminum chloride in solution? BT KEEP SMILING!!! UNASANA COLLEGE Page 11