Calculate the number of millilitre of `NH_(3)`(aq) solution (d=0.986g/ml) contain 2.5% by mass `NH_(3)`, which will be required to precipitate iron as `Fe(OH)_(3)` in a 0.8 g sample that contains 50% `Fe_(2)O_(3)`.

Options

A. 0.344 mL
B. 3.44 mL
C. 17.24 mL
D. 10.34 mL

Answer

D

Explanation

The precipitation reaction is
`Fe^(3+)+3NH_(3)+3H_(2)O to Fe(OH)_(3)+3NH_(4)^(+)`
mole of `Fe_(2)O_(3)` in sample =`(0.80 xx 0.5)/(160)=2.5 xx 10^(-3)`
mole of `Fe^(3+)=2 xx 2.5 xx 10^(-3)=5 xx 10^(-3)`
`M_(NH_(3))=(0.986 g // mL xx 1000 mL // "litre" xx 0.025)/(17)`
`implies M_(NH_(3))=1.45`
`3 xx "moles of" Fe^(3+)="moles of" NH_(3)`
`implies 1.45 xx V(in L)`
`V=(3 xx 5 xx 10^(-3))/(1.45)=10.35 xx 10^(-3) L` or 10.35 mL