Question

A mixture containing only FeCl3 and AlClz weighs 5.95 g . The chlorides are converted to the hydrous oxides and ignited to Fe2O3 and Al2O3 . The oxide mixture weighs 2.62 g . Calculate the percent Fe and Al in the original mixture .​

Answer 1

Answer:

The answer is "Al= 9.8% and Fe=18.0%"

Explanation:

Given:

The weight of $FeCl_3\ and \ ALCl_3$ = 5.95g

$gFeCl_3=gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})\\\\gAlCl_3=gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})$

$\to a = FeCl_3+AlCl_3\\\\\to a=x+y \\\\ \to a= 5.95$

$\to a =x \ gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})+ y \ gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})$

$\to x (\frac{162.2}{55.85})+ y (\frac{133.34}{26.98})= 5.95\\\\\to 2.90x+4.94y=5.95\\\\\ similarly \ for \ oxidies:\\\\\to 143x+1.89y=2.62\\\\\to x= 1.07 \ \ \ and \ \ y= 0.58\\\\\to \ Al \% = \frac{0.58}{5.95} \times 100= \bold{9.8} \%\\\\\to \ Fe \% = \frac{1.07}{5.95} \times 100= \bold{18.0} \%$