Stichiometry Chapter 1 (All SlO covered) PDF

Mass /mass relationship Q.NO.1 Calculate the number of gram of K2SO4 and water produced when 14g of KOH are reacted with excess of H2SO4. Also calculate the number of molecules of water produced: Equation: 2KOH(aq) + H2SO4(aq) → K2SO4(aq) +2H2O(l) Given data: mass of KOH= 14g formula mass of KOH =56...

Mass /mass relationship Q.NO.1 Calculate the number of gram of K2SO4 and water produced when 14g of KOH are reacted with excess of H2SO4. Also calculate the number of molecules of water produced: Equation: 2KOH(aq) + H2SO4(aq) → K2SO4(aq) +2H2O(l) Given data: mass of KOH= 14g formula mass of KOH =56g/mol Mass of K2SO4=? K2SO4= 𝟑𝟗𝑿𝟐 + 𝟑𝟐 + 𝟏𝟔𝒙𝟒 = 𝟏𝟕𝟒 g/mol Molecules of H2O =? 𝟏𝟒 Mole of KOH= = 0.25 𝟓𝟔 Balance equations shows that KOH = K2SO4 2moles KOH gives = 1mole K2SO4 𝟏 0.25 mole KOH = 𝒙 𝟎. 𝟐𝟓 of K2SO4 = 0.125moles 𝟐 Mass of K2SO4= n x formula mass of k2SO4 = 0.125x 174 = 21.75 g To find molecules of water, We know Number of particles = n x NA Compare mole of KOH verses mole of H2O Balance equation shows that KOH = H2O 2 moles KOH gives = 2 moles of water 0.25 moles KOH = 0.25 moles of H2O NUMBER OF MOLECULES OF WATER = n X NA = 0.25 X 6.02 X 1023 = 1.5 X 1023 molecules of water Q.No.2 magnesium metal reacts with HCl to give hydrogen gas. What is the minium volume of HCl solution (27% by weight) required to produce 12.1g of H2. The density of HCl is 1.14g/cm-3 Mg(s) + 2HCl(aq)→ MgCl2(aq) + H2(g) GIVEN DATA Mass of H2 = 12.1g 𝟐𝒈 Molar mass of H2= 𝒎𝒐𝒍 Volume of HCl= ? Density of HCl= 𝟏. 𝟏𝟒𝒈/𝒄𝒎 − 𝟑 𝟏𝟐.𝟏 Moles of H2 = 𝟐 = 𝟔 𝒎𝒐𝒍𝒆𝒔 Chemical equation shows that H2 = HCl 1 mole = 2 moles of HCl 6 moles = 12 moles of HCl Mass of HCl = n x molar mass of HCl = 12 x 36.5 g/mol = 438 g We know that 27 % by weight HCl mean 27 g HCl is present = in 100 g of solution 𝟏𝟎𝟎 1 g HCl = 𝟐𝟕 𝟏𝟎𝟎 438 g HCl present = 𝑿𝟒𝟑𝟖 = 𝟏𝟔𝟐𝟐. 𝟐𝒈 𝟐𝟕 Volume of HCl = d x mass = 1.14 x 1622.2 Mass volume relationship Q.No. 3 Calculate the volume of CO2 gas produced at standard temperature and pressure by the combustion of 20g of CH4 chemical equation CH4(g) + 2O2(g) CO2 + 2H2O We know that according to Avagardos 1 mole of any gas at STP has fixed volume has 22.4 dm-3 Given data : Mass of CH4 = 20 g Volume of CO2 = ? 𝟐𝟎 Moles of CH4 = 𝟏𝟒 = 𝟏. 𝟐𝟓 𝒎𝒐𝒍𝒆𝒔 Balance equation shows that CH4 = CO2 1 MOLES methane gives = 1 mole of CO2 1.25 mole CH4 gives = 1.25 moles of CO2 So volume of 1.25 moles of CO2 = 1.25 x 22.4 dm-3 = 28 dm-3 Q.No.3 calculate the volume of O2 gas at STP for the complete combustion of two moles of carbon disulphide (CS2). Calculate the volume of CO2 and SO2 produced also. Solution: CS2(g) + 3O2(g) → CO2(g) + 2SO2(g) Given data: Moles of CS2= 2 Mole of O2=? Volume of O2=? Volume of CO2=? Volume of SO2=? Chemical equation shows CS2 = O2 1mole CS2 reacts = 3 moles of oxygen 2 moles CS2 reacts = 3x2=6 moles of oxygen At STP 1 mole of O2 has volume =22.4 dm-3 6 moles of oxygen has volume at STP = 6x 22.4dm-3 = 134.4dm-3 Volume- volume relationship Q.4.Hydrogen sulfide burns in oxygen according to the equation. 2H2S(g) + 3O2(g) 2H2O(g) + 2SO2(g) CALCULATE the volume of O2 at STP required to burn 900dm-3 of H2S and also find the volume of SO2 gas produced BALANCE EQUATION SHOWS THAT H2S = O2 2dm of H2S reacts with = 3dm-3 of O2 -3 𝟑 900 dm-3 of H2S will reacts = 𝑿𝟗𝟎𝟎 dm-3 of O2 𝟐 = 1350 dm-3 of O2 Similarly H2S = O2 2dm-3 H2S produced = 2dm-3 of SO2 900 dm-3 of H2S will produced= 900 dm-3 of SO2 Slo# 1.2.4 Masses and volume of gas at RTP (24 dm-3 ) room temp and pressure Methane burns exothermically in the presence of free oxygen at room temperature and pressure (RTP)as shown below. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) If 25 moles of CH4 combust to give equal mole of CO2, then the volume of CO2 formed will be ? CH4 = CO2 25 mole of CH4 gives= 25 moles of CO2 WE KNOW At RTP 1 mole of any gas has fix volume = 24L (dm-3) So 25 moles of CO2 at RTP has volume = 25 x24 = 600 dm-3 SLO.# 1.2.5. solve problems based on stoichiometry using mole ratio as conversion factor: Mole ratio: Mole ratio mean the ratio of number of moles of reactant to product in a balanced chemical reaction. Example:1 Methanol burns according to the following reaction. 2CH3-OH + 3O2(g) 2CO2(g) + 4H2O(g) If 3.5moles of methanol are burned in oxygen , calculate i. How many moles of oxygen will consumed ii. How many moles of H2O will produced Moles of methanol =3.5 moles Moles of oxygen=? Solution: conversion factor = mole ratio = 3moles ofO2 2moles of CH3OH Desired quantity of oxygen= given quantity x conversion factor 𝟑 = 𝟑. 𝟓 𝑿 = 5.25 moles of O 𝟐 Molecular mass of C6H12O6 = 6X12+12x1+6x16 =180 g/mol 𝐚𝐭𝐨𝐦𝐢𝐜 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐂 % age of C = 𝐱 𝟏𝟎𝟎 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝 𝟕𝟐 = 𝑿 𝟏𝟎𝟎 = 𝟒𝟎% 𝟏𝟖𝟎 𝟏𝟐 % age of H= 𝟏𝟖𝟎 𝑿𝟏𝟎𝟎 = 𝟔. 𝟔% ⬚ 𝟗𝟔 % age of O = 𝑿𝟏𝟎𝟎 = 𝟓𝟑% 𝟏𝟖𝟎 Slo# 1.3.2 deduce empirical and molecular formula of compounds EMPIRICAL FORMULA: chemical formula which determine the combining ratio of atoms in a molecule called E.M MOLECULAR FORMULA: Chemical formula which determine the actual number and types of atoms present in a molecule or compound DETERMINATION OF EMPIRICAL FORMULA Following 5 steps involve for the determination of empirical formulae 1. Elements detection (experiments) 2. Masses of an elements(experimentally determined) 3. %age of an elements %𝒂𝒈𝒆 4. Mole ratios = 𝒂𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 5. Atomic ratios ( divide all ratio by least number) Q.No# 1.367g of an organic comound have elements C,H and O2, was combusted in a stream of air to yield 3.002g CO2 and 1.640g H2O. what is its Empirical formula? Empirical formula: Step1 𝑪: 𝑯: 𝑶 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝑶𝟐 𝟏 𝒎𝒐𝒍𝒆 𝒐𝒇𝑪 % 𝒐𝒇 𝑪 = 𝒙 𝐱 𝟏𝟎𝟎 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝟏𝒎𝒐𝒍𝒆 𝒐𝒇 𝑪𝑶𝟐 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐇𝟐𝐎 𝟏𝒎𝒐𝒍𝒆 𝒐𝒇 𝑯𝟐 % of H = 𝒙 𝑥100 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝟏 𝒎𝒐𝒍𝒆 𝒐𝒇 𝑯𝟐𝑶 𝟑.𝟎𝟎𝟐 𝟏𝟐 % 𝐨𝐟 𝐂 = 𝐱 𝐱 𝟏𝟎𝟎 = 𝟓𝟗. 𝟗1 𝟏.𝟑𝟔𝟕 𝟒𝟒 1.640 2 % of H = x x100 = 13.16 %. 1..367 18 % of Oxygen = 100 − [% of C + % of H] = 100 − [59.91 + 13.6]= 46.75 %𝒂𝒈𝒆 Step 3 Mole ratio = 𝐚𝐭𝐨𝐦𝐢𝐜 𝒎𝒂𝒔𝒔 C : H : O 59.91 13.16 46.75 C= H= = 12 1.008 16 4.99 ∶ 13.16 ∶ 2.92 Step 4 Simplest no Divide mole ratio by least one C : H : O 𝟒.𝟗𝟗 𝟏𝟑.𝟏𝟔 𝟐.𝟗𝟐 = 𝟏. 𝟕 = 𝟒. 𝟓 =𝟏 𝟐.𝟗𝟐 𝟐.𝟗𝟐 𝟐.𝟗𝟐 𝟐 ∶ 𝟓 ∶ 𝟏 Empirical formula C2H5O Q. The combustion analysis of an organic compound shows that it contains 65.44% C, 5.50% H and 20.06% O. if the molecular mass of compound is 110.15g/mol. Calculate Empirical and molecular formula E.F = C3H3O Molecular formula = n x Empirical formula 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒏= 𝑬𝒎𝒊𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 Molecular mass of compound = 110.15 g/mol Empirical formula mass =C3H3O (3x12+1.008x3+1x16)=55.05 g/mol 𝟏𝟏𝟎.𝟏𝟓 𝒏= =𝟐 𝟓𝟓.𝟎𝟓 Molecular formula = nx E.F = 2 x C3H3O = C6H6O2 SLO# 1.4.Excess and Limiting reagent Limiting Reagent: those reactant which produced the least amount of product is known as limiting reactant.it control the amount of product due to its smaller amount Example: Look at the balanced equation for the formation of water molecules 2H2 + O2 → 2H2O Balance equation shows that H2 O2  H2O 2moles(4g) 1mole(32g) 2moles(18g) I. 2moles(4g) hydrogen reacts with 1mole(32g) oxygen to produce 2 moles(18g) water II. Its mean hydrogen is in lesser amount as compared to oxygen III. So Hydrogen is the limiting reactant Oxygen is in excess amount hence called excess reagent Identification of limiting reactant: Following 3 steps are used to identify the limiting reactant 1. Calculate the number of moles from the given amount of reactant 2. Find the number of moles of product by using balance equation 3. Identify the reactant which produce the least amount of product as limiting reactant. Excess Reactant: “ The reactant left un-used or unreacted after completion of reaction is called “ Reactant in excess. ” Example: NH3 gas can be prepared by heating two solids NH4Cl and Ca(OH)2. If a mixture containing 100g of each solid is heated, then a. Calculate the number of grams of NH3 produced b. Calculate the excess amount of reagent left unreacted. 2NH4Cl(s)+ Ca(OH)2(s) → CaCl2 + 2NH3 + 2H2O 1. Calculate moles of reactants Mass of NH4Cl =100g mass of Ca(OH)2= 100g  Molar mass of NH4Cl= 53.5g/mol molar mass of Ca(OH)2=74g/mol 𝐦𝐚𝐬𝐬 𝟏𝟎𝟎 Moles of NH4Cl = = = 1.87 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝟓𝟑. 𝟓 100 moles of Ca(OH)2= =1.35 74 2. Compare moles of reactants with moles of products NH4Cl = NH3 2moles : 2moles 1 : 1 1.87 : 1.87moles ( NH3) : Ca(OH)2 = NH3 1 mole : 2moles 1.35 : 2x1.35 2.7moles (NH3) 3. Identify limiting reactant Since NH4Cl gives least amount of product(NH3) so NH4Cl is the Limiting reactant. While Ca(OH)2 is the excess reactant Mass of NH3 = moles X molar mass Mass of NH3= 𝟏. 𝟖𝟕𝒙 𝟏𝟕 = 𝟑𝟏. 𝟕𝟗 𝒈 b. amount of excess reactant left behind compare moles of reactant NH4Cl = Ca(OH)2 2 moles : 1mole 𝟏 1.87 : 𝑿 𝟏. 𝟖𝟕 = 𝟎. 𝟗𝟑𝟓 𝒎𝒐𝒍𝒆 𝟐 Hence the number of moles of Ca(OH)2 which completely reacts with 1.87 mole of NH4Cl is (0.935 mole) No of moles of Ca(OH)2 taken= 1.35 moles No of mole of Ca(OH)2 used= 0.935 No of moles of Ca(OH)2 left behind = 1.35-0.935 = 0.415 Mass of excess reactant Ca(OH)2left behind = moles x molar mass = 0.415 x 74 =30.71g Q.No. Calculate the number of gram of Al2S3 which can be prepared by the reaction of 20g of Al and 30g of sulphur. How much the non-limiting reactant is in excess? Reaction : 2Al + 3S → Al2S3 Solution: mass of Al = 20g mass of S = 30g Atomic mass of Al= 27 atomic mass of S = 32 𝐠𝐢𝐯𝐞𝐧 𝐦𝐚𝐬𝐬 𝟐𝟎 Moles of Al= = = 0.740 𝐚𝐭𝐨𝐦𝐢𝐜 𝐦𝐚𝐬𝐬 𝟐𝟕 𝟑𝟎 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐒 = = 𝟎. 𝟗𝟑𝟕 𝟑𝟐 C0mpare moles of reactant with Moles of prroducts Al = Al2S3 C 2 mole : 1mole 𝟏 0.740 : 𝑿 𝟎. 𝟕𝟒 = 𝟎. 𝟑𝟕 𝟐 S = Al2S3 Equation shows 3mole : 1 mole 𝟏 0.93 : 𝑿 𝟎. 𝟗𝟑 = 𝟎. 𝟐𝟖 moles of Al2S3 𝟑 Sulphur gives the least amount of product (Al2S3) so S is L.R Non-limiting (excess)reactant = Al Compare moles of reactants S = Al 3mole = 2mole 𝟐 0.93 moles S completely reacts with 𝟑 𝑿 𝟎. 𝟗𝟑 = 𝟎. 𝟔𝟐𝟒 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐀𝐥 Mole of Al taken = 0.937 Mole of Al used = 0.624 mole of Al unreacted = 0.937-0.624 = 0.313 mass of Al = 0.313x 27 =8.451g mass of Al2S3 = 0.28 x 150 = 42g Consider the reaction. 2Al + 6HBr → 2AlBr3 + 3H2 ( Past paper) If 6.44 moles of Al reacts with 9.92 moles of HBr to produce 4.96 moles of H2, then the number of moles of excess reactant left over at the end of reaction would be? Compare moles of reactant and product Al = H2 Equation shows 2 mole = 3moles 𝟑 6.44 moles give: 𝒙 𝟔. 𝟒𝟒 = 𝟗. 𝟔𝟔 𝐦𝐨𝐥𝐞𝐬 𝟐 HBr = H2 Equation shows 6moles = 3moles 𝟑 9.92 moles gives: 𝑿𝟗. 𝟗𝟐 = 𝟒. 𝟗𝟔 𝐦𝐨𝐥𝐞𝐬 𝟔 HBr is the limiting reactant because it gives the least amount of product Excess reactant = Al To find the moles of excess reactant compare reactants moles HBr = Al 6 : 2 moles 𝟐 9.92 : 𝟔 𝑿𝟗. 𝟗𝟐 = 𝟑. 𝟑𝟎 𝐦𝐨𝐥𝐞𝐬 Moles of Al taken = 6.44 Moles of Al used = 3.30 Moles of Al unreacted = 6.44-3.30 = 3.14 Q.NO. Aluminum reacts with bromine to form Aluminum bromide,as shown below 2Al + 3Br2 → 2AlBr3 if 15.8g of aluminum and 55.6g of bromine are available for reaction, then determine a. Limiting reactant b. Mass of AlBr3 produced.  Slo# 1.5.1 distinguish among theoretical yield, actual yield and percentage yield;(U) Actual yield Theoretical Percentage yield yield i. The amount of The amount of Chemist usually product which product interested in the are obtained in calculated efficiency of a chemical from balance reaction. which is reaction is chemical known as percentage known as actual equation is yield. yield. known as % 𝒚𝒊𝒆𝒍𝒅 ii. Actual yield theoretical 𝒂𝒄𝒕𝒖𝒂𝒍 𝒚𝒊𝒆𝒍𝒅 always lower yield. = 𝑿 𝟏𝟎𝟎 𝒕𝒉𝒆𝒐𝒓𝒂𝒕𝒊𝒄𝒂𝒍 𝒚𝒊𝒆𝒍𝒅 than Theoretical theoretical yield always yield greater than iii. Several reason actual yield. might possible a. Inexperience worker, side reactions b. If separation techniques like filtration, distillation and crystallization were not properly performed.so actual yield decrease  SLO# 1.5.2 calculate the percentage yield of a product in a chemical reaction Q.No. Silicon carbide (SiC) is an important ceramic material. It is produced by allowing sand (SiO2) to react with carbon at high temperature. SiO2 + 3C → SiC + 2CO When 100 kg sand is reacted with excess carbon, 51.4 kg of SiC is produced. What is the percentage yield of SiC? Balance equation shows that SiO2 = SiC 28+16x2=60 = 28+12 = 𝟒𝟎 60 g/mol sand = 40 g/mol of SiC. 𝟒𝟎 100 g sand will produce = X100 = 𝟔𝟔. 𝟔𝟔 𝒈 𝑺𝒊𝑪 𝟔𝟎 Actual yield of SiC = 51.4g Theoratical yield of SiC = 𝟔𝟔. 𝟔𝟔𝒈 𝐚𝐜𝐭𝐮𝐚𝐥 𝐲𝐢𝐞𝐥𝐝 % 𝐲𝐢𝐞𝐥𝐝 = 𝐗 𝟏𝟎𝟎 𝐭𝐡𝐞𝐨𝐫𝐚𝐭𝐢𝐜𝐚𝐥 𝐲𝐢𝐞𝐥𝐝 𝟓𝟏.𝟒 = 𝐱 100 = 𝟕𝟕. 𝟏𝟕% 𝟔𝟔.𝟔 Self-check exercise Q. No.a. How many moles of oxygen molecule are there in 50dm-3 of oxygen at STP b. what volume does 0.8 mole of N2 gas occupy at STP? c. calculate the number of molecules of O2 produced by the thermal decomposition of 490g of KClO3? Thank you Prepared by Sir Ejaz Ali Lecturer AKHSS,K