Question

A coffee cup calorimeter with a heat capacity of 6.80 JrC was used to measure the change in enthalpy of a precipitation reaction. A 50.0 ml. solution of 0.360 M AgNO, was mixed with 50.0 ml. of 0.500M KSCN. After mixing, the temperature was observed to increase by 4.03 C Calculate the enthalpy of reaction, AH per mole of precipitate formed (AgSCN). Assume the specific heat of the product solution is 4.14J/(g * C) and that the density of both the reactant solutions is 1.00 gml. Calculate the theoretical moles of precipitate formed from AgNO, and KSCN moles of precipitate formed from AgNO, mol moles of precipitate formed from KSCN mol Calculate the heat change experienced by the calorimeter contents, qt 4conts Calculate the heat change expicrenced by the calorimcter contents, 4 4al Calculate the hecat change produced by the solution process, qui 4ion Calulate AHi for one mole of precipitate formed. AH kJ/mole

Answer 1

Ccal = 6.50 5/°c 8T = 4:03°c Csolutions = 4.143199 Total volume (V) = 50+50 = loome density (d) = loog me mass of solution (m) = dxu =loog/mexloome Run = 100g Ag Novet KSCN CALAg SCN CS) + KAQ (aq) 1. imot imol imol im.moly of Ag No = mtu -0.360 X SO 0.018mol o As Agnoz is limiting reagent = 18 mmol motes of Agsen formed = 18 mual from AgNO3 = 1:8x102 mol q=ms AT - q - longa . 3 mols of Agson formed from ksen - 0.500 M x 0.050L = 0.025 mol ♡ Ycontents Embat 1 = 100g x 24:145/90 cх4.03°C Scanned with CamScanner -1668.42): i

q cal = Cal AT = 6-805/06 x 4.03°c = 27.4045 Isolution = quae + & contents = 27.404 + 1668.42 = 1668.42J 1.6765 Att Solution =-q = -1668.625 0.018 mol --92690 J/mol -- 92.7 kJ/mol Scanned with CamScanner