The amount of water produced (in g) in the ox | vedclass

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(C) The oxidation of rhombic sulphur $(S_8)$ by concentrated $HNO_3$ produces sulphuric acid $(H_2SO_4)$ as the product with the highest oxidation sta

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$288$

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(C) The oxidation of rhombic sulphur $(S_8)$ by concentrated $HNO_3$ produces sulphuric acid $(H_2SO_4)$ as the product with the highest oxidation state of sulphur $(+6)$.The balanced chemical equation is:$S_8 + 48 \ HNO_3 \longrightarrow 8 \ H_2SO_4 + 48 \ NO_2 + 16 \ H_2O$From the stoichiometry of the reaction,$1$ mole of $S_8$ produces $16$ moles of $H_2O$.Since $1$ mole of rhombic sulphur is $S_8$,we must consider the reaction for $1$ mole of $S_8$ atoms (as implied by the question context of $1$ mole of sulphur atoms,or if $1$ mole of $S_8$ is meant,the stoichiometry adjusts accordingly. Given the standard interpretation of $1$ mole of $S$ atoms):For $1$ mole of $S$ atoms,the equation is: $S + 6 \ HNO_3 \longrightarrow H_2SO_4 + 6 \ NO_2 + 2 \ H_2O$.Thus,$1$ mole of $S$ produces $2$ moles of $H_2O$.Mass of $H_2O = 2 \ mol 	imes 18 \ g \ mol^{-1} = 36 \ g$.However,if the question refers to $1$ mole of $S_8$ molecules,then $16$ moles of $H_2O$ are produced.Mass of $H_2O = 16 \ mol 	imes 18 \ g \ mol^{-1} = 288 \ g$.

The amount of water produced (in $g$) in the oxidation of $1$ mole of rhombic sulphur by conc. $HNO_3$ to a compound with the highest oxidation state of sulphur is. . . . . (Given data : Molar mass of water $= 18 \ g \ mol^{-1}$)

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