Question

A 1.26-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of the chloride is recovered as 3.26 g of silver chloride. Identify the metal.

Answer 1

Answer:

The metal is calcium (to form CaCl2)

Explanation:

Step 1: Data given

Mass of sample of an alkaline earth metal chloride = 1.26 grams

Silvernitrate (AgNO3) = in excess

All of the chloride is recovered as 3.26 g of silver chloride.

Step 2: The equation

XCl2 + 2AgNO3 → X(NO3)2+ 2 AgCl

Step 3: Calculate moles AgCl

Moles AgCl = mass AgCl / molar mass AgCl

Moles AgCl = 3.26 grams / 143.32 g/mol

Moles AgCl = 0.0227 moles

Step 4: Calculate moles XCl2

For 1 mol XCl2 we'll have 2 moles AgCl

For 2*0.0227 moles we need 0.0227/2 = 0.01135 moles

Step 5: Calculate molar mass XCl2

Molar mass XCl2 = mass / moles XCl2

Molar mass XCl2 = 1.26 grams / 0.01135 moles

Molar mass XCl2 = 111 g/mol

Step 6: Calculate of metal X

Molar mass X = molar mass XCl2 - molar mass Cl2

Molar mass X = 111 - 70.9 = 40.1 g/mol

The metal is calcium (to form CaCl2)