Consider the following reactions. `2Fe_(2) S_(3)(s) + 6 H_(2) O(l)+3 " O"_(2)(g) rarr 4" Fe"(OH)_(3)(s)+6S` The number of moles of `Fe_(2) S_(3)` are 2, `H_(2)O` is 2 and 3 moles of `O_(2)` to react. Then find no. of moles of `Fe(OH)_3` are

Options

A. `2.62`
B. `3.62`
C. `1.33`
D. `2.43`

Answer

C

Explanation

`underset(2 " mol")(2Fe_(2) S_(3)) (s) +underset(6" mol") (6 H_(2) O (l)) + underset( 3" mol")(3O_(2) (g)) rarr underset(4" mol")(4 Fe (OH)_(3)(s)) + 6S`
1 mol `Fe_(2) S_(3)` form = 2 mol `Fe (OH)_^3)`
2 mol `H_(2))` form `=4/6 xx 2 = 1.33` mol `Fe (OH)_(3)`
3 mol `O_(2)` form = 4 mol `Fe(OH)_(3)`
Thus, `H_(2)O` is the limiting reactant
and `Fe(OH)_(3)` formed is `=1.33` mol