The minimum amount of $\mathrm{O}_{2}(\mathrm{~g})$ consumed per gram of reactant is for the reaction : (Given atomic mass: $\mathrm{Fe}=56, \mathrm{O}=16, \mathrm{Mg}=24, \mathrm{P}=31, \mathrm{C}=12, \mathrm{H}=1$ )

(C) $4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})$

(A) $\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

(B) $\mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{~s})$

(D) $2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{~s})$

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Some Basic Concepts of Chemistry

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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

)

(A) $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 CO_{2} (g) + 4 H_{2} O (l)$
(B) $P_{4} (s) + 5 O_{2} (g) \to P_{4} O_{10} (s)$
(C) $4 Fe (s) + 3 O_{2} (g) \to 2 Fe_{2} O_{3} (s)$ ✓ Correct
(D) $2 Mg (s) + O_{2} (g) \to 2 MgO (s)$

Answer: (C)

Solution

Compare the oxygen used per gram of fuel. Propane burns five

O X_{2}

to its

44 g

mole,

160/44 \approx 3.6 g

of oxygen per gram. White phosphorus uses five

O X_{2}

per

124 g

, about

1.3 g

per gram. Magnesium uses one

O X_{2}

for two moles,

32/48 \approx 0.67 g

per gram. Iron needs three

O X_{2}

for four moles,

96/224 \approx 0.43 g

per gram - the smallest of all, so iron consumes the least oxygen per gram.

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Some Basic Concepts of Chemistry

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