A) 240 g
B) 480 g
C) 720g
D) 960g
Correct Answer: B
Solution :
\[3C+\frac{3}{2}{{O}_{2}}\xrightarrow[{}]{{}}3CO\] \[F{{e}_{2}}{{O}_{3}}+3CO\xrightarrow[{}]{{}}2Fe+3C{{O}_{2}}\] 1 mole of\[F{{e}_{2}}{{O}_{3}}\equiv 3\]mole of\[CO\equiv \frac{3}{2}\]mole of\[{{O}_{2}}\] \[\because \]160 g of\[F{{e}_{2}}{{O}_{3}}\]require \[{{O}_{2}}=\frac{3}{2}\times 32=48g\] \[\therefore \]1.60 kg of\[F{{e}_{2}}{{O}_{3}}\]will require \[{{O}_{2}}=480\text{ }g\]
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