QJEEM19Ch55a8: JEE Main 2019 Chemistry - Class 11 JEEM19Ch55a8 Solution: The minimum amount of O 2 (g) consumed per gram of reactant is…

(a) 4Fe(s) + 3O₂(g) $\to$ 2Fe₂O₃(s)

${{Moles\,of\,{O_2}} \over 3} = {{Moles\,of\,{Fe}} \over 4}$

$\Rightarrow$ Moles of O₂ = ${3 \over 4}$ $\times$ ${1 \over {56}}$ moles = ${1 \over {24.8}}$ moles = ${3 \over {224}} \times 32$ g of O₂ = 0.43 g of O₂

(b) P₄(s) + 5O₂(g) $\to$ P₄O₁₀(s)

${{Moles\,of\,{O_2}} \over 5} = {{Moles\,of\,{P_4}} \over 1}$

$\Rightarrow$ Moles of O₂ = 5 $\times$ ${1 \over {124}}$ moles = ${1 \over {24.8}}$ moles = ${1 \over {24.8}} \times 32$ g of O₂ = 1.3 g of O₂

(c) C₃H₈(g) + 5O₂(g) $\to$ 3CO₂(g) + 4H₂O(l)

${{Moles\,of\,{O_2}} \over 5} = {{Moles\,of\,{C_3}{H_8}} \over 1}$

$\Rightarrow$ Moles of O₂ = 5 $\times$ ${1 \over {44}}$ moles = ${1 \over {24.8}}$ moles = ${1 \over {8.8}} \times 32$ g of O₂ = 3.6 g of O₂

(d) 2Mg(s) + O₂(g) $\to$ 2MgO(s)

${{Moles\,of\,{O_2}} \over 1} = {{Moles\,of\,{Mg}} \over 2}$

$\Rightarrow$ Moles of O₂ = ${1 \over 2}$ $\times$ ${1 \over {24}}$ moles = ${1 \over {48}}$ moles = ${1 \over {48}} \times 32$ g of O₂ = 0.66 g of O₂