Mixture of 10 moles of `Fe_(2)S_(3), 20` moles of `H_(2)O` and 30 moles of `O_(2)` react with `5%` yield in the given reaction :
`Fe_(2)S_(3) + H_(2)O + O_(2) rarr Fe(OH)_(3) + S`
Then moles of `Fe(OH)_(3)` that can be produced is -

Options

A. `(10)/(3)`
B. `(20)/(3)`
C. `20`
D. 10

Answer

B

Explanation

`{:(Fe_(2)S_(3),+,3H_(2)O,+(3)/(2)O_(2),overset(50%)(rarr),2Fe(OH)_(3),+,3S,),(10 " mole ",,20 " mole",,30 " moles",,,,):}`H_(2)O` is the limiting reagent.
moles of `Fe(OH)_(3) = (2)/(3) xx 20 xx 0.5 = (20)/(3)` moles