Question

I’m just checking my answer. Thanks!

Answer 1

78.89 g of CO is required.

Step-by-step explanation:

`Fe_(2)O_(3)+3CO\longrightarrow3CO_(2)+2Fe`

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
`Fe_(2)O_(3)`.

2 moles of Fe is produced on reacting 3 moles of CO.

`Weight\:of\:one\:mole\:of\:Fe_(2)O_(3)=2*55.85+3*15.99=159.67g\\`

`Weight\:of\:one\:mole\:of\:CO=12.01+15.99=28g`

`The\:weight\:of\:two\:moles\:of\:Fe=2*55.85=111.7g\\111.7g\:of\:Fe\:is\:produced\:by\:reacting\:(3*28=)84g\:of\:CO\\\\104.9g\:of\:Fe\:is\:produced\:on\:reacting=(84*104.9)/(111.7)=78.89g\:of\:CO\\`

Hence, 78.89 g of CO on reacting with excess of ferric oxide will produce 104.9 g of Fe.