Question

A 0.750-g sample of a compound that might be iron(II) chloride, nickel(II) chloride, or zinc chloride is dissolved in water, and 22.40 mL of 0.515 M AgNO3 is required to completely precipitate all the chloride ion as AgCl. Name the compound, and write its formula.

Answer 1

Answer: The compound is nickel (II) chloride.

NiCl2

Explanation:

Let's assume that the compound is X.

Then,

XCl2 + 2AgNO3 ---> X(NO3)2 + 2AgCl

Given,

Volume of AgNO3 = 22.4ml = 0.0224L

Concentration of AgNO3 = 0.515M

no of moles = concentration x volume

:• n(AgNO3) = 0.515 x 0.0224

n(AgNO3) = 0.011536 mol

Therefore, no of moles of AgNO3 is 0.0011536mol

From the equation above, we can deduce that;

n(XCl2) = 1/2n(AgNO3)

:• n(XCl2) = 1/2 × 0.011536

n(XCl2) = 0.005768

Therefore number of moles of XCl2 is 0.005768 mol.

Mass, m, of XCl2 = 0.75g (Given)

Molar mass = mass/ no of moles

Molar mass of XCl2 = 0.75/0.005768

:• molar mass of XCl2= 130.03g/mol

Molar mass of Cl = 35.453

Cl2= 2 x 35.453 = 70.906

Molar mass of XCl2 = M.mass of X +

M.mass of Cl2

:• 130.03 = M.mass of X + 70.906

Molar mass of X = 130.03 - 70.906

Molar mass of X = 59.1

By comparing to literature, we can conclude that X is Nickle ,Ni.

Therefore, the compound is NiCl2 , Nickle(II) Chloride.