1.

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 -

Answer»

`(10)/(3)`
`(20)/(3)`
`20`
10

Solution :`{:(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


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