JEE Main 2019 — Chemistry Physical Chemistry

2019mcqmedium

The minimum amount of ${O}_{2}(g)$ consumed per gram of reactant is for the reaction:

(Given atomic mass: $Fe=56, O=16, Mg=24, P=31, C=12,H=1$ )

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

(i) 5 mole ${O}_{2}$ required = 1 mole of ${P}_{4}$ =124 gm ${P}_{4}$

for per gram, ${O}_{2}$ required = $\frac{5}{124} mole$

(ii) 1 mole ${O}_{2}$ required = 2 mole Mg = $2\times 24 gm$

for per gram, ${O}_{2}$ required $=\frac{1}{48} mole$

(iii) 3 mole ${O}_{2}$ required = 2 mole Fe = $4\times 56 gm$

for per gram Fe, ${O}_{2}$ required $=\frac{3}{112\times 2}$ 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 $=\frac{5}{44}$ mole

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JEE Main Physical Chemistry in other years

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JEE Main 2019 — Chemistry Physical Chemistry

2019mcqmedium

The minimum amount of ${O}_{2}(g)$ consumed per gram of reactant is for the reaction:

(Given atomic mass: $Fe=56, O=16, Mg=24, P=31, C=12,H=1$ )

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

(i) 5 mole ${O}_{2}$ required = 1 mole of ${P}_{4}$ =124 gm ${P}_{4}$

for per gram, ${O}_{2}$ required = $\frac{5}{124} mole$

(ii) 1 mole ${O}_{2}$ required = 2 mole Mg = $2\times 24 gm$

for per gram, ${O}_{2}$ required $=\frac{1}{48} mole$

(iii) 3 mole ${O}_{2}$ required = 2 mole Fe = $4\times 56 gm$

for per gram Fe, ${O}_{2}$ required $=\frac{3}{112\times 2}$ 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 $=\frac{5}{44}$ mole

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2019 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry