Answer:
78.89 g of CO is required.
Explanation:
[tex]Fe_{2}O_{3}+3CO\longrightarrow3CO_{2}+2Fe[/tex]
Fe = 55.85 g/mol
O = 15.99 g/mol
C = 12.01 g/mol
2 moles of Fe is produced on reacting 1 mole of [tex]Fe_{2}O_{3}[/tex].
2 moles of Fe is produced on reacting 3 moles of CO.
[tex]Weight\:of\:one\:mole\:of\:Fe_{2}O_{3}=2\times55.85+3\times15.99=159.67g\\[/tex]
[tex]Weight\:of\:one\:mole\:of\:CO=12.01+15.99=28g[/tex]
[tex]The\:weight\:of\:two\:moles\:of\:Fe=2\times55.85=111.7g\\111.7g\:of\:Fe\:is\:produced\:by\:reacting\:(3\times28=)84g\:of\:CO\\\\104.9g\:of\:Fe\:is\:produced\:on\:reacting=\frac{84\times104.9}{111.7}=78.89g\:of\:CO\\[/tex]
Hence, 78.89 g of CO on reacting with excess of ferric oxide will produce 104.9 g of Fe.
