Ich versuche, das folgende LaTeX mithilfe der vimtexkontinuierlichen Kompilierung zu kompilieren vim:
Latex:
\documentclass[12pt]{article}
\title{Electrolysis of Water and Aqueous Solutions}
\author{Carlo Abelli \\ AP Chemistry --- Mr.\ Kern}
\date{February 19, 2015}
\usepackage[margin=1.0in]{geometry}
\begin{document}
\begin{titlepage}
\maketitle
\thispagestyle{empty}
\end{titlepage}
\section{Water Electrolysis Calculations and Questions}
\subsection{Half Reaction for Each Electrode}
Cathode (reduction): \(2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)}\)\\
This reaction took place at the cathode because during the splint test the gas
produced a popping sound, characteristic of hydrogen, a highly flammable gas.
\\
\\
Anode (oxidation): \(2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} +
4e^{-}\)\\
This reaction took place at the anode because during the splint test the gas
reignited the splint ember, characteristic of oxygen.
\subsection{Overall Reaction}
\(2H_{2}O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)}\)
\subsection{Average Current}
Average Current: \(\frac{0.44 + 0.43 + 0.44 + 0.44 + 0.45 + 0.46 + 0.45 + 0.46
+ 0.46 + 0.45 + 0.45 + 0.46 + 0.45 + 0.47 + 0.46}{15} = 0.45\)
\subsection{Amount of Hydrogen Produced in One Hour}
\(\frac{0.45\ coulombs}{1.00\ \sec} * \frac{60.0\ \sec}{1.00\ \min} *
\frac{60.0\ \min}{1.00\ hour} * \frac{1.00\ e^{-}}{1.60*10^{-19}\ coulombs} *
\frac{1.00\ mol\ e^{-}}{6.022*10^{23}\ e^{-}} * \frac{1.00\ mol\ H_{2}}
{2.00\ mol\ e^{-}} = 0.0084\ mol\ H_{2}\)\\
\(0.0084\ mol\ H_{2} * 22.4\ L/mol = 0.19\ L\ H_{2}\)\\
\(0.0084\ mol\ H_{2} * 2.016\ g/mol = 0.0017\ g\ H_{2}\)
\subsection{Amount of Time Needed to Produce 30 mL Oxygen Gas at
22\(^{\circ}\)C and 1.00 atm}
\(1.00\ atm * 0.030\ L = n * 0.08206\ L\ atm\ K^{-1}\ mol^{-1} * 295\ K\)\\
\(n = 0.0012\ mol\ O_{2}\)\\
\(0.0012\ mol\ O_{2} * \frac{4.00\ mol\ e^{-}}{1.00\ mol\ O_{2}} *
\frac{6.022*10^{23}\ e^{-}}{1.00\ mol\ e^{-}} *
\frac{1.60*10^{-19}\ coulombs}{1\ e^{-}} * \frac{1.00\ \sec}{0.45\ coulombs}
= 100.\ \sec\)
\subsection{Moles of Gas Produced in Experiment}
Cathode (Hydrogen)\\
\(P_{container} = 36.8\ cm\ H_{2}O * \frac{10.0\ mm}{1.00\ cm} *
\frac{1.08\ g\ H_{2}O} {1\ mL\ H_{2}O} * \frac{1\ mL\ Hg}
{13.534\ g\ Hg} + 751.1\ mm\ Hg = 780.5\ mm\ Hg\)\\
\(P_{gas} = P_{container} - P_{H_{2}O}\)\\
\(P_{gas} = 780.5\ mm\ Hg - 18.7\ mm\ Hg = 761.8\ mm\ Hg\)\\
\(761.8\ mm\ Hg * 0.0486\ L = n * 62.363\ L\ mmHg\ K^{-1}\ mol^{-1} * 294.0\ K\)
\\
\(n = 0.00202\ mol\ H_{2}\)\\
\\
Anode (Oxygen)\\
\(P_{container} = 24.0\ cm\ H_{2}O * \frac{10.0\ mm}{1.00\ cm} *
\frac{1.08\ g\ H_{2}O} {1\ mL\ H_{2}O} * \frac{1\ mL\ Hg}
{13.534\ g\ Hg} + 751.1\ mm\ Hg = 770.3\ mm\ Hg\)\\
\(P_{gas} = P_{container} - P_{H_{2}O}\)\\
\(P_{gas} = 770.3\ mm\ Hg - 18.7\ mm\ Hg = 751.6\ mm\ Hg\)\\
\(751.6\ mm\ Hg * 0.0239\ L = n * 62.363\ L\ mmHg\ K^{-1}\ mol^{-1} * 294.0\ K\)
\\
\(n = 0.000980\ mol\ O_{2}\)
\subsection{Theoretical Moles of Gas Produced}
Cathode (Hydrogen)\\
\(\frac{0.45\ coulombs}{1.00\ \sec} * \frac{60.0\ \sec}{1.00\ \min} * 14.0\ \min *
\frac{1.00\ e^{-}}{1.60*10^{-19}\ coulombs} * \frac{1.00\ mol\ e^{-}}
{6.022*10^{23}\ e^{-}} * \frac{1.00\ mol\ H_{2}}{2.00\ mol\ e^{-}} =
0.0020\ mol\ H_{2}\)\\
\\
Anode (Oxygen)\\
\(\frac{0.45\ coulombs}{1.00\ \sec} * \frac{60.0\ \sec}{1.00\ \min} * 14.0\ \min *
\frac{1.00\ e^{-}}{1.60*10^{-19}\ coulombs} * \frac{1.00\ mol\ e^{-}}
{6.022*10^{23}\ e^{-}} * \frac{1.00\ mol\ O_{2}}{4.00\ mol\ e^{-}} =
0.00098\ mol\ O_{2}\)
\subsection{Percent Deviation}
Cathode (Hydrogen)\\
\(\frac{0.00202\ mol - 0.0020\ mol}{0.0020\ mol} * 100 = 0.0\%\ deviation\)\\
\\
Anode (Oxygen)\\
\(\frac{0.000980\ mol - 0.00098\ mol}{0.00098\ mol} * 100 = 0.0\%\ deviation\)
\section{Aqueous KI, KBr and KCl Electrolysis Calculations and Questions}
\subsection{Half Reactions for Each Electrode}
\begin{tabular}{ l r @{\(\rightarrow\)} l }
KI Solution\\
Cathode (reduction): & \(2H_{2}O_{(l)} + 2e^{-}\) &
\(2OH^{-}_{(aq)} + H_{2(g)}\)\\
Anode (oxidation): & \(2I^{-}_{(aq)}\) & \(I_{2(s)} + 2e^{-}\)\\
\\
KBr Solution\\
Cathode (reduction): & \(2H_{2}O_{(l)} + 2e^{-}\) &
\(2OH^{-}_{(aq)} + H_{2(g)}\)\\
Anode (oxidation): & \(2Br^{-}_{(aq)}\) & \(Br_{2(l)} + 2e^{-}\)\\
\\
KCl Solution\\
Cathode (reduction): & \(2H_{2}O_{(l)} + 2e^{-}\) &
\(2OH^{-}_{(aq)} + H_{2(g)}\)\\
Anode (oxidation): & \(2Cl^{-}_{(aq)}\) & \(Cl_{2(g)} + 2e^{-}\)\\
\end{tabular}\\
\\
The water half reaction had to have occurred at the cathode because the
phenolphthalein indicator was activated by the creation of \(OH^{-}\),
creating the purple color observed at the cathode.\\
\\
The iodine, chlorine and bromine half reactions had to have occurred at the
anode because in the first experiment, a dark solid was created at the anode
(iodine), in the second experiment, a yellow liquid was observed (bromine)
and in the third experiment, a chlorine smelling gas was produced (chlorine).
\subsection{Overall Equations}
\begin{tabular}{ l r @{\(\rightarrow\)} l }
KI Solution: & \(KI_{(aq)} + 2H_{2}O_{(l)}\) & \(KOH_{(aq)} + 2H_{2(g)} +
I_{2(s)}\)\\
KBr Solution: & \(KBr_{(aq)} + 2H_{2}O_{(l)}\) & \(KOH_{(aq)} + 2H_{2(g)} +
Br_{2(l)}\)\\
KCl Solution: & \(KCl_{(aq)} + 2H_{2}O_{(l)}\) & \(KOH_{(aq)} + 2H_{2(g)} +
Cl_{2(g)}\)\\
Test Solution:
\end{tabular}
\subsection{Half Reactions and Overall Equation for Molten KI, KBr and KCl}
KI\\
\begin{tabular}{ l r @{\(\rightarrow\)} l }
Cathode (reduction): & \(K^{+}_{(l)} + e^{-}\) & \(K_{(s)}\)\\
Anode (oxidation): & \(2I^{-}_{(l)}\) & \(I_{2(s)} + 2e^{-}\)\\
Overall Equation: & \(2KI_{(l)}\) & \(2K_{(s)} + I_{2(s)}\)\\
\end{tabular}\\
\\
KBr\\
\begin{tabular}{ l r @{\(\rightarrow\)} l }
Cathode (reduction): & \(K^{+}_{(l)} + e^{-}\) & \(K_{(s)}\)\\
Anode (oxidation): & \(2Br^{-}_{(l)}\) & \(Br_{2(l)} + 2e^{-}\)\\
Overall Equation: & \(2KBr_{(l)}\) & \(2K_{(s)} + Br_{2(s)}\)\\
\end{tabular}\\
\\
KCl\\
\begin{tabular}{ l r @{\(\rightarrow\)} l }
Cathode (reduction): & \(K^{+}_{(l)} + e^{-}\) & \(K_{(s)}\)\\
Anode (oxidation): & \(2Cl^{-}_{(l)}\) & \(Cl_{2(g)} + 2e^{-}\)\\
Overall Equation: & \(2KCl_{(l)}\) & \(2K_{(s)} + Cl_{2(g)}\)\\
\end{tabular}
\subsection{Time Required to Produce 1.25 g Iodine at 0.75 amps}
\(1.25\ g\ I_{2} * \frac{1.00\ mol\ I_{2}}{253.80\ g} *
\frac{2.00\ mol\ e^{-}}{1.00\ mol\ I_{2}} *
\frac{6.022*10^{23}\ e^{-}}{1.00\ mol\ e^{-}} *
\frac{1.60*10^{-19}\ coulombs}{1.00\ e^{-}} *
\frac{1.00\ \sec}{0.75\ coulombs} *
\frac{1.00\ \min}{60.0\ \sec} = 21\ \min\)
\subsection{Time Required to Produce 0.100 L Hydrogen gas in Lab Conditions at
0.75 amps}
\(751.1\ mmHg * 0.100\ L\ H_{2} = n * 62.363\ L\ mmHg\ K^{-1}\ mol^{-1} *
294\ K\)\\
\(n = 0.00410\ mol\)\\
\(0.00410\ mol\ H_{2} * \frac{2.00\ mol\ e^{-}}{1.00\ mol\ H_{2}} *
\frac{6.022*10^{23}\ e^{-}}{1.00\ mol\ e^{-}} *
\frac{1.60*10^{-19}\ coulombs}{1.00\ e^{-}} *
\frac{1.00\ \sec}{0.75\ coulombs} * \frac{1.00\ \min}{60.0\ \sec} = 18\ \min\)
\end{document}
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Warum passiert das? Ich kann es nicht herausfinden. Wenn Sie es pdflatexüber die Befehlszeile ausführen, wird tatsächlich das PDF generiert (nachdem einige Warnungen ausgegeben wurden), also muss das Problem von oder herrühren latexmk, richtig?vimtexvim
Antwort1
Das Problem wurde mithilfe \usepackage{fixltx2e}von make \(und \)robust behoben. Kompiliert jetzt ohne Fehler.