How much volume of 5 M HNO_3 is required to | vedclass

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Question

(B) The balanced redox reaction is: $3Fe^{+2} + NO_3^- + 4H^+ \rightarrow 3Fe^{+3} + NO + 2H_2O$. From the stoichiometry,$3$ moles of $Fe^{+2}$ react

Vedclass · Accepted Answer

$19.05$

Vedclass · Answer

(B) The balanced redox reaction is: $3Fe^{+2} + NO_3^- + 4H^+ ightarrow 3Fe^{+3} + NO + 2H_2O$. From the stoichiometry,$3$ moles of $Fe^{+2}$ react with $1$ mole of $HNO_3$. Moles of $Fe^{+2} = \frac{16 \ g}{56 \ g/mol} = 0.2857 \ mol$. Moles of $HNO_3$ required = $\frac{1}{3} 	imes 0.2857 \ mol = 0.09524 \ mol$. Volume of $HNO_3$ $(V)$ = $\frac{	ext{moles}}{	ext{Molarity}} = \frac{0.09524 \ mol}{5 \ M} = 0.01905 \ L$. Converting to $mL$: $0.01905 \ L 	imes 1000 \ mL/L = 19.05 \ mL$.

How much volume of $5 \ M$ $HNO_3$ is required to oxidize $16 \ g$ of $Fe^{+2}$ if $HNO_3$ is converted into $NO$? (Answer in $mL$)

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