Answer:
The answer is "[tex]41.23 \ L\ N_2[/tex]"
Explanation:
[tex]2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049 \frac{\ g NO_3^{-}}{mol})} \times \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\[/tex]
[tex]=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23[/tex]
The volume of CO2 is 206.27 L
The ideal gas equation is used to determine the volume, pressure, temperature, or number of moles. It can be mathematically expressed as:
PV = nRT
From the given information:
The equation for the reaction can be expressed as:
[tex]\mathbf{2NO_3^-_{(aq)} + 5CO_{(g)} + 2H^+_{(aq)} \to N_2{(g)} + 5CO_2_{(g)} + H_2O_{(l)}}[/tex]
The number of moles of CO2 from the reaction is;
[tex]\mathbf{= \dfrac{215 \ g}{62.0049} \times \dfrac{5 \ mol \ of \ CO_2}{2 \ mol \ of \ NO_3^-} }[/tex]
[tex]\mathbf{= 8.669 \ moles \ of \ CO_2 }[/tex]
From ideal gas, by making the volume the subject of the, we have:
The volume of CO₂ [tex]\mathbf{V= \dfrac{nRT}{P}}[/tex]
[tex]\mathbf{V= \dfrac{8.669 \ moles \times 0.08205 L atm/ kmol \times 290\ K}{1 \ atm }}[/tex]
[tex]\mathbf{V= 206.27 \ L \ of \ CO_2 \ gas}[/tex]
Learn more about the volume of CO2 gas here 206.27 L
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