For the given series of reactions:NH_{3(g)} + | vedclass

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(D) The balanced chemical equations are:$4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_2O_{(l)}$$4 NO_{(g)} + 2 O_{2(g)} \rightarrow 4 NO_{2(g

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$17/56$

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(D) The balanced chemical equations are:$4 NH_{3(g)} + 5 O_{2(g)} ightarrow 4 NO_{(g)} + 6 H_2O_{(l)}$$4 NO_{(g)} + 2 O_{2(g)} ightarrow 4 NO_{2(g)}$Adding these equations,we get the overall reaction:$4 NH_{3(g)} + 7 O_{2(g)} ightarrow 4 NO_{2(g)} + 6 H_2O_{(l)}$To obtain the maximum mass of $NO_2$,the reactants must be in their stoichiometric ratio to ensure there is no limiting reagent.The molar ratio is $n_{NH_3} : n_{O_2} = 4 : 7$.The mass ratio is calculated as:$	ext{Mass ratio} = (4 	imes 	ext{molar mass of } NH_3) : (7 	imes 	ext{molar mass of } O_2)$$	ext{Mass ratio} = (4 	imes 17) : (7 	imes 32)$$	ext{Mass ratio} = 68 : 224$Dividing both sides by $4$,we get $17 : 56$.

For the given series of reactions:
$NH_{3(g)} + O_{2(g)} \rightarrow NO_{(g)} + H_2O_{(l)}$
$NO_{(g)} + O_{2(g)} \rightarrow NO_{2(g)}$
To obtain the maximum mass of $NO_2$ from a given mass of a mixture of $NH_3$ and $O_2$,the ratio of the mass of $NH_3$ to $O_2$ should be:

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For the given series of reactions:$NH_{3(g)} + O_{2(g)} \rightarrow NO_{(g)} + H_2O_{(l)}$$NO_{(g)} + O_{2(g)} \rightarrow NO_{2(g)}$To obtain the maximum mass of $NO_2$ from a given mass of a mixture of $NH_3$ and $O_2$,the ratio of the mass of $NH_3$ to $O_2$ should be: