Question

1. How many moles of oxygen are required to burn 2.4 moles of ethane (C.H.). 2. How much Barium Sulfate is produced when 3.18 g of Barium Nitrate react with 4.6 g of sodium sulfate. Ba(NO3)2 + Na SO BaSO. + NaNO 3. Calculate the mass of Aluminum Chloride that can be produced by the reaction of 129 g of Al with 106 g Cl.

Answer 1

solution: - ( Given - Cato Burning of etherne give cat H₂O thon- CzH6 +702 ->200, +217,0 Here we see 1 mole of burning of ethare require 7 thone require 3.5 mol 2.4 moles of ethome require = 3.5x214 = 8.4 moles of O2 Given : - Ba(NO3), + Na, 504 - B9 504 + Na (NO) Aller bleme inag reaction - Ba(NO3)2 + Naz 507 — BaSO4 + 2. Nacinos) molar mass 261 When I mole of each reach reuet of Ba(NO₃)₂ + Na₂SO4 will give 1 mole Basoa. So we have 3.18 gram Ba (NO), Then MolesGrivon weight = 3:18 -0.0122 moles And, 4.69 Sodium Scalphate 1. Then moles = 4.6 = 0.032 142 Here Sodium scelphate in excess cimount so Ba(N3)2 is limitting reagert — So Amount of Baso depend on Ba(NO3)2 . So moles of Basoy = 0.0122 Moles so weight = .0122 X Molar mass = 0.01224233 = 2.84 g Anger.

Given - 7 201 4342 2A143 Here 2 moles of Al react with 3 moles of cle to - give Here limitting reagent is Als because available of low amount so t mole Al react with us to give a Gien- 1299 AL Fen mole - 129 -4.70 moles of Al 27 & 106g Cl₂ then Moles will given by = Given weight inolarmas = 106 = 1.49 Moles El 71 here Cl2 is in less amount so product it, and this will be limitting reagent. Now we see depend on 15 cl required = .: 1049 1 mol Al X149 = 0.993 mole Ali So, mass of Al= 0.9934 Molar mass = 0.993 X27 = 26.814 97