The minimum amount of \(\mathrm{O_2(g)}\) consumed per gram of reactant is for the reaction :
(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
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- \(\mathrm{4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)}\)
- \(\mathrm{P_4(s)+5O_2(g)\rightarrow P_4O_{10}(s)}\)
- \(\mathrm{C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)}\)
- \(\mathrm{2Mg(s)+O_2(g)\rightarrow 2MgO(s)}\)
The Correct Option is A
Solution and Explanation
(A) \(\dfrac{3\times32}{4\times56}=\dfrac{96}{224}=0.43\)
(B) \(\dfrac{5\times32}{124}=\dfrac{160}{124}=1.29\)
(C) \(\dfrac{5\times32}{44}=\dfrac{160}{44}=3.64\)
(D) \(\dfrac{32}{2\times24}=\dfrac{32}{48}=0.67\)
The minimum is for reaction (A).
Answer: (1) \(\mathrm{4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)}\)
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