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

Aluminum

Step-by-step 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}* 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.