Question

139%

Answer 1

Answer:

The percent composition of this compound is 94%

Explanation:

The reaction can be formed as

$2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}$

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

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

$\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}$

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

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

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%