MCQ
One and a half mole of oxygen combine with aluminium to form $Al_2O_3$, then the weight of aluminium metal used in this reaction is .............. $\mathrm{gms}$ (Atomic weight of $Al = 27$)
- A$27$
- B$81$
- C$108$
- ✓$54$
$1$ moles of oxygen react with $108 / 3\, g$ of $Al$
And $0.5$ mole of oxygen react with $108 \times 0.5/ 3\, g$ of $Al$
$\mathrm{Al}=18$ $gram$
$1$ moles of oxygen react with $108 / 3\, g$ of $Al$
And $1$ mole of oxygen react with $108 \times 1/ 3\, g$ of $Al$
$\mathrm{Al}=36$ $gram$
$=18+36$
$54\,gm$
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The correct decreasing order of enthalpies of reaction for producing carbocation is


$H _{2} F _{2( g )} \rightarrow H _{2( g )}+ F _{2( g )}$
$\Delta U =-59.6\,kJ\,mol ^{-1} \text { at } 27^{\circ}\,C$.
The enthalpy change for the above reaction is (-) $kJ mol ^{-1}$ [nearest integer] Given : $R =8.314\,JK ^{-1}\,mol ^{-1}$