A) 3 : 2
B) 2 : 1
C) 1 : 2
D) 1 : 1
Correct Answer: A
Solution :
For first oxide, Moles of oxygen\[=\frac{22}{16}=1.375,\] Moles of\[Fe=\frac{78}{56}=1.392\] Simpler molar ratio, \[\frac{1.375}{1.375}=1,\frac{1.392}{1.375}=1\] \[\therefore \]The formula of first oxide is\[FeO\]. Similarly for second oxide, Moles of oxygen\[=\frac{30}{16}=1.875,\] Moles of\[Fe=\frac{70}{56}=1.25\] Simpler molar ratio\[=\frac{1.875}{1.25}=1.5,\frac{1.25}{1.25}=1\] \[\therefore \]The formula of second oxide is\[F{{e}_{2}}{{O}_{3}}\]. Suppose in both the oxides, iron reacts with \[x\text{ }g\]oxygen. \[\therefore \]Equivalent weight of Fe in\[FeO\] or \[=\frac{weight\,of\,F{{e}_{II}}}{weight\,of\,oxygen}\times 8\] \[\frac{56}{2}=\frac{weight\,of\,F{{e}_{II}}}{x}\times 8\] ?. (i) \[\therefore \]Equivalent weight of Fe in \[F{{e}_{2}}{{O}_{3}}=\frac{weight\,of\,F{{e}_{II}}}{weight\,of\,oxygen}\times 8\] \[\frac{56}{3}=\frac{weight\,of\,F{{e}_{II}}}{x}\times 8\] ?. (ii) From Eq. (i) and (ii), \[\frac{weight\,of\,F{{e}_{II}}}{weight\,of\,F{{e}_{III}}}=\frac{3}{2}\]
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