8Al + 30HNO₃ → 8 Al(NO₃)₃+3NH₄NO₃ + 9H₂O as per thegiven equation the number of moles of the aluminium metal that can be oxidised by one mole of HNO₃ is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

3/8

30/8

8/3

8/30

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

8Al + 30HNO₃ → 8 Al(NO₃)₃+3NH₄NO₃ + 9H₂O;
Al → Al⁺³ (↑3)
$\large ^{ + 5}NO_3^ - \; \to {\;^{ - 3}}NH_4^ + ( \downarrow 8)$
Of the 30 moles $\large NO_3^ -$ ions 27 moles of nitrate ions remained unchanged only 3 mole nitrate ions reduced to $\large NH_4^ +$ ions
All the 8mole of 'Al' oxidized to Al⁺³
So 8 moles of Al reduces 3 moles of nitrate ion
'X' moles of Al reduces 3 moles of nitrate ion
$\large X\;=\;\frac{3}{8}$
Alternate method
$\large \begin{array}{*{20}{c}} {\mathop {Al}\limits_O \; \to \;\mathop {A{l^{ + 3}}}\limits_3 } \\ { increase \;in\;oxidation\;number\; = \;3} \end{array}\left| {\begin{array}{*{20}{c}} {\mathop {^{ + 5}NO_3^ - }\limits_{ + 5} \to \mathop {N{H_4^{+}}}\limits_{ -3} } \\ {decrease \;in\;oxidation\;numbe\; = \;8} \end{array}} \right.$
Criss - Cross method
8 mole of Al reduces 3moles of $\large NO_3^ -$ ions
'X' mole of Al reduces 1moles of $\large NO_3^ -$ ions
$\large X\;=\;\frac{3}{8}$