Question

I’m just checking my answer. Thanks!

Answer 1

Answer:

78.89 g of CO is required.

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.

$Weight\:of\:one\:mole\:of\:Fe_{2}O_{3}=2\times55.85+3\times15.99=159.67g$

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

$The\:weight\:of\:two\:moles\:of\:Fe=2\times55.85=111.7g\\111.7g\:of\:Fe\:is\:produced\:by\:reacting\:(3\times28=)84g\:of\:CO\\\\104.9g\:of\:Fe\:is\:produced\:on\:reacting=\frac{84\times104.9}{111.7}=78.89g\:of\:CO$

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