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The minimum amount of $O_{2}\left ( g \right )$ consumed per gram of reactant is for the reaction :

(Given atomic mass : $Fe=56,$ $O=16$ , $Mg=24,P=31,C=12,H=1$ )
Option 1)
$4Fe\left ( s \right )+3O_{2}\left ( g \right )\rightarrow 2Fe_{2}O_{3}\left ( s \right )$

Option 2)
$P_{4}\left ( s \right )+5O_{2}\left ( g \right )\rightarrow P_{4}O_{10}\left ( s \right )$
Option 3)
$C_{3}H_{8}\left ( g \right )+5O_{2}\left ( g \right )\rightarrow 3CO_{2}\left ( g \right )+4H_{2}O\left ( I \right )$
Option 4)
$2Mg\left ( s \right )+O_{2}\left ( g \right )\rightarrow 2MgO\left ( s \right )$

Answers (1)

best_answer

$3$ mole $O_{2}$ required for $4$ mole $f_{e}$ $4\times fe=4\times 56=224$

per gram $fe$ , $O_{2}$ required $=\frac{5}{224}mole$ (minimum)

per gram $P_{n},O_{2}$ required= $\frac{5}{124}mole$

per gram $C_{3}H3,O_{2}$ required= $\frac{5}{44}mole$

per gram $M_{g},O_{2}$ required = $\frac{1}{48}mole$

Option 1)
$4Fe\left ( s \right )+3O_{2}\left ( g \right )\rightarrow 2Fe_{2}O_{3}\left ( s \right )$

Option 2)

$P_{4}\left ( s \right )+5O_{2}\left ( g \right )\rightarrow P_{4}O_{10}\left ( s \right )$

Option 3)

$C_{3}H_{8}\left ( g \right )+5O_{2}\left ( g \right )\rightarrow 3CO_{2}\left ( g \right )+4H_{2}O\left ( I \right )$

Option 4)

$2Mg\left ( s \right )+O_{2}\left ( g \right )\rightarrow 2MgO\left ( s \right )$

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