0
answers
0
watching
149
views
27 Nov 2019

Alkaline I2 (FW = 253.8089) reacts with warfarin, C19H16O4 (FW = 308.3297), to form one mole of iodoform, CHI3 (FW = 393.7322), for each mole of the parent compound reacted. The liberated iodoform can be analyzed by:CHI3 + 3Ag+ + H2O -> 3AgI(s) + 3H+ + CO(g)In one experiment, the CHI3 liberated from 12.3932 g sample was treated with 25.00 mL of 0.02628 M AgNO3 (FW = 169.8732), and the excess Ag+ (FW = 107.8682) was titrated with 3.34 mL of 0.05073 M KSCN (FW = 97.1813). Calculate the percentage warfarin in this sample.Ag+ + SCN- -> AgSCN(s)

For unlimited access to Homework Help, a Homework+ subscription is required.

discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Weekly leaderboard

Start filling in the gaps now
Log in