Question

A compound of formula XCl3 reacts with aqueous AgNO3 to yield solid AgCl according to the following equation: XCl3(aq)+3AgNO3(aq)→X(NO3)3(aq)+3AgCl(s) When a solution containing 0.521 g of XCl3 was allowed to react with an excess of aqueous AgNO3, 1.68 g of solid AgCl was formed. What is the identity of the atom X?

Answer 1

Answer:

Aluminum

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

Mass of $AgCl$ = 1.68 g

Molar mass of $AgCl$ = 143.32 g/mol

Thus,

$Moles\ of\ AgCl=\frac {1.68}{169.87}=0.01172$

From the reaction below:

$XCl_3_{(aq)}+3AgNO_3_{(aq)}\rightarrow X(NO_3)_3_{(aq)}+3AgCl_{(s)}$

3 moles of $AgCl$ are produced when 1 mole of $XCl_3$ undergoes reaction.

So,

1 mole of $AgCl$ are produced when $\frac {1}{3}$ mole of $XCl_3$ undergoes reaction.

0.01172 mole of $AgCl$ are produced when $\frac {1}{3}\times 0.01172$ mole of $XCl_3$ undergoes reaction.

Thus, moles of $XCl_3$ = 0.0039 moles

Let the atomic mass of X = x g/mol

atomic mass of chlorine = 35.5 g/mol

Thus, Molar mass of $XCl_3$ = x + 3(35.5) g/mol = x + 106.5 g/mol

Moles = 0.0039 moles

Mass = 0.521 g

Thus, molar mass = Given mass/ Moles = 0.521 / 0.0039 = 133.5897 g/mol

So,

x + 106.5 = 133.5897

x = 27.0897 g/mol

This Atomic weight corresponds to Aluminum. Hence, X is aluminum.