A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?

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Annainn Annainn · Answer 1 · 2019-08-13T15:04:39+02:00

Answer:

Mass% chloride ion = 29.32%

Explanation:

Given:

Mass of alkali metal chloride sample = 2.5624 g

Mass of the precipitate = 3.03707 g

To determine:

The mass percent of chloride ion in the sample

Calculation:

The reaction of alkali metal with silver nitrate results in the precipitation of the chloride ion as silver chloride (AgCl).

Molecular weight of AgCl =143.32 g/mol

[tex]Moles(AgCl)=\frac{Mass(AgCl)}{Mol.wt(AgCl)}=\frac{3.03707g}{143.32g/mol}=0.02119moles[/tex]

Since 1 mole of AgCl contains 1 mole of Cl, therefore:

# moles of Cl = 0.02119 moles

At wt of Cl = 35.45 g/mol

[tex]Mass(Chloride)=moles*at.wt = 0.02119moles*34.45g/mol=0.7512g[/tex]

[tex]Mass%(chloride)=\frac{mass(chloride)}{mass(sample)}*100=\frac{0.7512}{2.5624}*100 = 29.32%[/tex]