Question

Answe all please!

Consider the following reaction carried out under constant pressure 2HCl(aq) + Zn(s) - H2(g) + ZnCl2(s) AHpx = -80.97 kJ Calculate the heat associated with the complete reaction of 24.3 g of HCl with 16.8 g of Zn. 0-54.0 kJ 0-10.8 kJ O-55.2 kJ -20.8 kJ O-27.0 kJ

Answer 1

2 HU aq + Zn o, H2 19g + 2nd2; Ah = - 80.97 kg 72-92 g 65.38g Given: 24:39 16.89 AH=2 65-38 ΥΛΟ Limiting reactant: . 65.38g an reach completely with 72.92 g HU - 16.8g zu will react with = 72-92 x 16.8 g Hy = 18.7379 HU but given mars af HU = 24-39 . Zn is dinitry reactant Theoretical yield of heat : 65.389 zu completely react to produce 80.9719 heart 2. 16.89 žu will react to produce = 80.97x16=8 IeJ heruf = 20.81 kJ heat e IV 65-38 . Answer: - 20.8 kJ

?о? ои Co23) + 2 Hio — си зғ 2Pa 8 * о се уч з н-опу, это • 10-е. Athron in terms of Bond energees is given by AMaryn = {nx BE pactants - EnxBe products - (1186-о * изве», -(«чве +2+36 ) - (2x744 + 4* *** ) - (х ча4 + 2х41 в) = 34Sч — 26 52 = Qo 2 ) ". Апту : Фо14) :: Дmsuw to : Lo2 к. СО