JEE Main 201910 AprilEvening Shift

JEE Main 2019 · 10 April · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

The minimum amount of O _ 2 (g) consumed per gram of reactant is...

The minimum amount of

{O}_{2}\left(g\right)

consumed per gram of reactant is for the reaction:

(Given atomic mass:

Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1

)

Options

A ${P}_{4}\left(s\right) + 5{O}_{2}\left(g\right) \to {P}_{4}{O}_{10}\left(s\right)$
B $2Mg\left(s\right) + {O}_{2}\left(g\right) \to 2MgO\left(s\right)$
C $4Fe\left(s\right) + 3{O}_{2}\left(g\right) \to 2F{e}_{2}{O}_{3}\left(s\right)$
D ${C}_{3}{H}_{8}\left(g\right) + 5{O}_{2}\left(g\right) \to 3C{O}_{2}\left(g\right) + 4{H}_{2}O ( l )$

C. $4Fe\left(s\right) + 3{O}_{2}\left(g\right) \to 2F{e}_{2}{O}_{3}\left(s\right)$

(i) 5 mole

{O}_{2}

required = 1 mole of

{P}_{4}

=124 gm

{P}_{4}

for per gram,

{O}_{2}

required =

\dfrac{5}{124}mole

(ii) 1 mole

{O}_{2}

required = 2 mole Mg =

2 \times 24gm

for per gram,

{O}_{2}

required

= \dfrac{1}{48}mole

(iii) 3 mole

{O}_{2}

required = 2 mole Fe =

4 \times 56gm

for per gram Fe,

{O}_{2}

required

= \dfrac{3}{112 \times 2}

mole

(iv) 5 mole

{O}_{2}

required -

= 1

mole

{C}_{3}{H}_{8}

=44 gm

{C}_{3}{H}_{8}

per gram

{C}_{3}{H}_{8}, {O}_{2}

required

= \dfrac{5}{44}

mole