$(a)\;21.5\%,78.5\%\qquad(b)\;22\%,78\%\qquad(c)\;20.25\%,79.75\%\qquad(d)\;20\%,80\%$

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Let the initial wt of the mixture is 100gm then the final wt of the mixture after heating in air will be 105gm.

Let $x$ be the wt of $FeO$ in the initial mixture then wt of $Fe_3O_4=100-x$

When the mixture is heated in air.$(O_2)$

$4FeO+O_2\rightarrow 2Fe_2O_3$

$4Fe_3O_4+O_2\rightarrow 6Fe_2O_3$

4 moles $FeO$=2 moles of $Fe_2O_3$

$\Rightarrow \large\frac{x}{72}$ moles=$\large\frac{x}{144}$ moles of $Fe_2O_3$

4 moles $Fe_2O_4$=6 moles of $Fe_2O_3$

$\Rightarrow \large\frac{100-x}{232}$ moles=$\large\frac{6}{4}\big(\frac{100-x}{232}\big)$ moles of $Fe_2O_3$

Total moles of $Fe_2O_3=\large\frac{x}{144}+\frac{6}{4}\big(\large\frac{100-x}{232}\big)$

Wt of $Fe_2O_3=\big(\large\frac{x}{144}+\frac{6}{4}\big(\frac{100-x}{232})]$$\times 160=105$

$x=20.25gm$=wt of $FeO$

$\Rightarrow $ Wt of $Fe_3O_4=100-x$=79.75gm

% FeO=20.25

$\%Fe_3O_4$=79.75$

Hence (c) is the correct answer.

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