JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 9)
The minimum amount of O2(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)
4Fe(s) + 3O2(g) $$ \to $$ 2Fe2O3(s)
P4(s) + 5O2(g) $$ \to $$ P4O10(s)
C3H8(g) + 5O2(g) $$ \to $$ 3CO2(g) + 4H2O(l)
2Mg(s) + O2(g) $$ \to $$ 2MgO(s)
Explanation
(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
$${{Moles\,of\,{O_2}} \over 3} = {{Moles\,of\,{Fe}} \over 4}$$
$$ \Rightarrow $$ 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
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