Question

A 0.732 g sample of a compound that might be iron(II) chloride, nickel(II) chloride, calcium 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 silver chloride. What is the compound?

Answer 1

Answer:

Molar Mass of X = 56 g/mol

56 g/mol is the Molar Mass of iron (Fe), So X is iron (Fe)

Reaction will become:

$FeCl_2+2AgNO_3 \rightarrow Fe(NO_3)_2+2AgCl$

Explanation:

Consider the reaction:

$XCl_2+2AgNO_3 \rightarrow X(NO_3)_2+2AgCl$

Volume of $AgNO_3$=22.40 ml=0.0224 L

concentration of $AgNO_3$=0.515 M

Moles of $AgNO_3$=n= Volume of

Moles of $AgNO_3$=n= 0.224 * 0.515

Moles of $AgNO_3$=n= 0.011536 mol

1 Mole of $XCl_2$ reacts with 2 moles of  $AgNO_3$

Moles of  $XCl_2$ = $\frac{Moles\ of\ AgCl_2 }{2}$

Moles of $XCl_2$=0.011536/2

Moles of $XCl_2$= 0.05768 mol

Mass of $XCl_2$= 0.732 g

Molar Mass of $XCl_2$=Mass/Number of moles

Molar Mass of $XCl_2$= 0.732/0.05768

Molar Mass of $XCl_2$=126.9 g/mol

Now,

Molar Mass of $Cl_2=70.9 g/mol$

Molar Mass of $XCl_2$ = Molar Mass of X + Molar Mass of $Cl_2$

126.9=Molar Mass of X + 70.9

Molar Mass of X = 56 g/mol

56 g/mol is the Molar Mass of iron (Fe), So X is iron (Fe)

Reaction will become:

$FeCl_2+2AgNO_3 \rightarrow Fe(NO_3)_2+2AgCl$

Answer 2

As sample of 0.732 g of a compound that may be iron and chloride and nickel including the calcium and chloride or the zinc is mixed in H2O and has a 22.40 ml  0.515 M Ag No3 is needed to complete precipitate.

Hence the output will be

• Molar Mass of X = 56 g/mol is the Molar Mass of iron (Fe), So X is iron (Fe)
• Volume of =22.40 ml=0.0224 L
• concentration of =0.515 M
• Moles of =n= Volume of
• Moles of =n= 0.224 * 0.515, Moles of =n= 0.011536 mol.

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