To obtain the maximum mass of NO_2 from a giv | vedclass

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(C) The balanced chemical equation for the reaction is: $4 NH_3 + 7 O_2 \longrightarrow 4 NO_2 + 6 H_2O$ To obtain the maximum product without any lim

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$\frac{17}{56}$

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(C) The balanced chemical equation for the reaction is: $4 NH_3 + 7 O_2 \longrightarrow 4 NO_2 + 6 H_2O$ To obtain the maximum product without any limiting reagent,the reactants must be in their stoichiometric ratio. The molar ratio of $NH_3$ to $O_2$ is $4:7$. The mass ratio is calculated as: $\frac{	ext{Mass of } NH_3}{	ext{Mass of } O_2} = \frac{4 	imes 	ext{Molar mass of } NH_3}{7 	imes 	ext{Molar mass of } O_2}$ $= \frac{4 	imes 17}{7 	imes 32} = \frac{68}{224} = \frac{17}{56}$

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 based on the reaction: $NH_3 + O_2 \longrightarrow NO_2 + H_2O$

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