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

(Given atomic mass :Fe=56,O=16Mg=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)

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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