Question

an ore of iron contains FeS and some non volatile impurity. on roasting if this ore converts all FeS into Fe2O3 and 8% loss in weight was observed. calculate the mass percentage of FeS in the ore?​

Answer 1

Explanation:

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Answer 2

Given:

Iron ore containing $FeS$ and impurity on roasting convert the $Fe_{2}O_{3}$  into  and 8% loss in weight observed

To Find:

Mass% of in the iron ore = ?

Solution:

Let Weight of iron ore = 100

Weight of $FeS$  = x

Weight of impurity = y

Weight of iron ore is equal to weight of $FeS$   and impurity

100 = x  +  y → first equation

Moles of $FeS$ =  $\dfrac{x}{88}$

Weight of  $Fe_{2}O_{3}$  = $\dfrac{x 160}{88 }$× 2

                =  0.909x

8% loss in weight when all   $FeS$ converted to  $Fe_{2}O_{3}$

0.909x+y=96 y  → second equation

B solving first & second equation

x = $\dfrac{4}{0.091}$

x = 43.95

Percentage of  $FeS$ = 44%