Addition of AgNO3 to aqueous solutions of the complex results in a cloudy white precipitate, presumably AgCl. You dissolve 0.1000 g of the complex in H2O and perform a precipitation titration with 0.0500 M AgNO3 as the titrant. Using an electrode that is sensitive to [Ag ], you reach the endpoint after 9.00 mL of titrant is added. How many grams of chloride ion were present in the 0.1000-g sample

0.016 grams of chloride ion were present in the 0.1000 grams of sample.

Explanation:

According to question, 9.00 mL of titrant was added to solution with 0.1000 grams of complex to reach the end point.

Molarity of the silver nitrate solution = 0.0500 M

Volume of the silver nitrate solution = V = 9.00 mL = 0.009 L

1 mL = 1000 L

Moles of silver nitrate = n

n = 0.00045 mol

According to 1 mole of silver nitrate reacts with 1 mol of chloride ion, then 0.00045 moles of silver nitrate will :

of chloride ions

Mass of chloride ions :

0.00045 mol × 35.5 g/mol = 0.016 g

0.016 grams of chloride ion were present in the 0.1000 grams of sample.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g