Question

139%

A 3.56g sample of iron powder was heated in gaseous chlorine, and 10.39g of an iron
chloride was formed. What is the percent composition of this compound?
​

Answer 1

The percent composition of this compound is 94%

Step-by-step explanation:

The reaction can be formed as

`2 \mathrm{Fe}+3 \mathrm{Cl}_(2) \rightarrow 2 \mathrm{FeCl}_(3)`

`\frac{\text { Weight of } \mathrm{Cl}_(2)}{\text { 3* Molar Mass of } \mathrm{Cl}_(2)}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}`

`\frac{\text { Weight of } \mathrm{Cl}_(2)}{3 *(2 * 35.5)}=(3.56)/(2 * 55.8)`

`\text { Weight of } C l_(2)=(3.56 * 3 * 71)/(2 * 55.8)=6.79 \mathrm{g}`

`\mathrm{n}\left(\mathrm{Cl}_(2)\right)=\mathrm{m}\left(\mathrm{Cl}_(2)\right) / \mathrm{M}\left(\mathrm{Cl}_(2)\right)=6.79 / 71=0.1 \mathrm{m}`

`\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}`

Based on no. of iron reacted,

`\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_(3)\right)`

n = m/M

`\mathrm{m}\left(\mathrm{FeCl}_(3)\right)=\mathrm{n}^(*) \mathrm{M}=0.06^(*) 162.5=9.75 \mathrm{g}`

% composition of
`FeCl_3`

=
`(9.75 / 10.39)^(*) 100`

= 94%