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Examil Module4-5rx and soln v1 - Requires Respondus LockDown Browser Webcam: CHEM-1311-SY1/SY2 red 017 pts Question 13 According to the following reaction, what mass of PbCl2 can form from 235 mL of 0.110 M KCl solution? Assume that there is excess Pb(NO3)2 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq) 7.199 1.80 g 5.94 g 1.30 g 3.59 g

Answer 1

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Answer & 2 KClaq) + Pb(NO3)2 - PbCl2 (6) + 2 KNO3 (aq) ). 235 mL of 0.110M KCl solution, number of moles of Kal=n= 0.110*235 = 0.026 moles 1000 KCL. one mole of From reaction, a moles of kal yields PbCl2 Therefore, 0.026 moles kcl will yield 1 (0.026) moles = 0.013 of Pbc2 mass of PbCl2 = (no. of moles of PbCl2)* molar mass of Pb ) (278-2) g mass of PbCl2 = (0.013) mol mass of PbCl2 = 3.59 g mass of PbCl = 3.6 grams: Answer o