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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 -A. `(10)/(3)`B. `(20)/(3)`C. `20`D. 10 |
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Answer» Correct Answer - B `{:(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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