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 \[\text{FeO}\text{.}\] Similarly for second oxide, Moles of oxygen \[\text{=}\frac{30}{16}=1.875,\] 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 \[\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{.}\] Suppose in both the oxides, iron reacts with \[x\,g\]oxygen. \[\therefore \]Equivalent weight of Fe in \[\text{FeO}\] or \[\text{=}\,\frac{\text{weight}\,\text{of}\,\text{F}{{\text{e}}_{\text{II}}}}{\text{weight}\,\text{of}\,\text{oxygen}}\,\,\text{ }\!\!\times\!\!\text{ }\,\text{8}\] \[\frac{56}{2}=\frac{\text{weight}\,\text{of}\,\text{F}{{\text{e}}_{\text{II}}}}{x}\,\times \,8\] \[\therefore \]Equivalent weight of Fe in \[F{{e}_{2}}{{O}_{2}}=\frac{\text{weight}\,\text{of}\,\text{F}{{\text{e}}_{\text{III}}}}{\text{weight}\,\text{of}\,\text{oxygen}}\times \,8\] \[\frac{56}{3}=\frac{\text{weight}\,\text{of}\,\text{F}{{\text{e}}_{\text{III}}}}{x}\times \,\,8\] ?(ii) From Eq. (i) and (ii), \[\frac{\text{weight}\,\text{of}\,\text{F}{{\text{e}}_{\text{II}}}}{\text{weight}\,\text{of}\,\text{F}{{\text{e}}_{\text{II}}}}\,\text{=}\,\frac{\text{3}}{\text{2}}\]
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