# Find the mass of AgI in the precipitate. [Solved] Add the two expressions for NaI used and set the sum equal to the total moles of NaI:
(0.00425944 z) (0.103873 0.00440142 z) = 0.102
Solve for z algebraically:
z = 13.2 g AgI

Let z be the mass of AgI (in grams) to be found.
Then 23.6-z is the mass of HgI2.

Ag{ } NaI AgI Na{ }
z / (234.77267 g AgI/mol) x (1 mol NaI / 1 mol AgI) = (0.00425944 z) mol NaI used for the AgI

Hg{2 } 2 NaI HgI2 2 Na{ }
(23.6-z) / (454.3989 g HgI2/mol) x (2 mol NaI / 1 mol Hgi2) = (0.103873 0.00440142 z) mol NaI used for the HgI2

Add the two expressions for NaI used and set the sum equal to the total moles of NaI:
(0.00425944 z) (0.103873 0.00440142 z) = 0.102
Solve for z algebraically:
z = 13.2 g AgI

Add the two expressions for NaI used and set the sum equal to the total moles of NaI:
(0.00425944 z) (0.103873 0.00440142 z) = 0.102
Solve for z algebraically:
z = 13.2 g AgI

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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