Question

Please answer questions from Q1 to Q5. Thank you.

Solve the following questions; show all your work details. NO credits will be added if short answers are submitted. Q1) Automobiles air bags are inflated by nitrogen gas generated by the rapid decomposition of sodium azide, NaN3: NaN3 (8) ► Na(s) + N2 © (NOT balanced) If an air bag has a volume of 50.0 L and is to be filled with nitrogen gas at 0.828 atm and 26°C, how many grams of NaN3, must be decomposed? 5 points Q2) In a more recent developed process, ethanol can be prepared by reacting ethylene (found in petroleum) with water C2H4 (g) + H2O(g) → CH5OH) (NOT balanced) If 100.8 L of ethylene measured @ STP reacts with water vapor. It is found that only 186.3 g of ethanol is produced, what is the percent yield of the reaction? 5 points Q3) what is the density of isopropylene gas measured at pressure of 0.75 atm and temperature of 250°C? 5 points Q4) Hydrogen is generated when the reaction between zinc and hydrochloric acid takes a place according to the following chemical equation: Zn(s) + HCl → ZnCl2(aq) + H2(g) (NOT balanced) 1- How many kilograms of zinc were needed to produce 21.0 m² of H2 gas measured at a temperature of 22°C and a pressure of 1.5 atm? 2- Calculate the molarity of the hydrochloric acid used if the volume of the acid used was 0.260121 m. 10 points 05) Identify of a cycloalkane liquid which its vapor that has a density of 2.7 g/l measured at pressure of 0.9843 atm and temperature of 100°C. 5 points

Answer 1

Balanced sol 2 Na Nz (s) 2Na(s) TN2 (g) To find mass of NaNz We will use Ideal Gas Equation PV = nRT to calculate it. Given. Volume of air bag = So litres Pressure = 0.828 atm Temperature = 26°C = 299K (26+273 We have to take Gas constant R = 0.082 Latmk mol PV = n RT on being no. of moles n= PV of Ny gas RT

DOMS" Page No Date n= 0.828 X5o = 41.4 0.082 X 299 24.518 n= 1.69 moles 2 Naſz (s) 2Na(s) 12 (9) According to the reaction Imole of NaNg must produce a mole of Na and 3 mole of Nz gos. 13 e. 1.69 moles of Nagas must have been Produced by 21.69 moles of NaNg. no of moles = given mass (of a substance) Los Molecular Mass givenmass = no. of moles X Molecular Mass 1089 X 65 Xoo 67 = B.23 grams Ane - Amodant of NaNg decomposeda Holecular u 73-231 Balanced C Hy (g) + H₂O(g) → GHG OH (9) To find: Percent yield of reaction Given: volume of Thylene at STP = 100.8 litres AAST: Temperature =0°C = 273k Pressure = latm. mass of ethanol actually produced 186.39 for percentage geld, using Ideal Gas Equation?

BOMS Page No. Date / PV = nRT o n= PV n= I x 100.8 (R = 0·082 Latm] 0·082 X 273 = 4.5 moles of Ethanolgos estimated in= filen mars of ethosulf or produced maps here) Molar mass of Ethanol given mass of ethanol = nx Molor Maes Molar Mars =405 X 46 of thonod=46/ = 207g Estimated production of elhanol = 207g Actually produced ethanol = 186.39 percentage Vield = Petual mass y loo Theoretical Mass - Cor Estimated) = 186031100 = 90% 207 Ane Percentage Vield of Ethanol = 90% 3 To calculate density of isopropylene gos. Given pressure of 0.75 atm and temperature of 250°C (473k) o We have PV = nRT P=nRT Sole n = no of moles n= m M m M p = m MXV mass of a substance Molar mass of That substance RT

[ Isopropylene GHG] GoMS Page No. Date I I P = PRT M Holax pas L = PXM RT 2m = f (density) y = 0·75x42 = 0.812 g/L 0·082x473 Density of Isopropylene gas is 0.812 g/L mol R0.082 Latmk Aus * Zn (s) tice → Zn Cl + H₂ (g) To find : 0 Mass of Zinc (in kg) @ Molarity of Hydrochlorie aced [1m² = 10 Litres] Voleeme of H₂ (gas) = 21 m T = (21 x103) Litres Temperature =22Ć Pressure = 1.5 atm Ideal Cas Equation, PV = nRT n= PV - 1.5x 21X103 = 103 x10 moles . RT 0·082 X 295 = 2953 Using I mole of a gas must have been produced by I mole of Zn. . 9.3x10 moles of the must have been produced by 1.3410 moles of Zn a) n (of Zn) = 1.3410 - n=m T T m (given mass) = .nXM

POMS Page No. Date = 260 (21_ 16.112 M 16. 137 The - Molarity of Hell is 16-112 M To identify : a cycloalkane Given density = 2.79 / 4 Pressure = 0.9843 atm. Temperature = 100 °C = 373K 1 We will Ideal Gas Equation of the form stated in problem (3 as ! P= f RT I being density By calculating Molae mass of the compound, we will identity cycloalkane M = LRT О2p3 4 = 207 r0.082 X373 = 83.89 ~ 84g As cyclohexane; GH12 has Molar mass of og si Answer - Cycls hexane) ů The Cycloalkane