☰

将一个小页面与另一个拉伸的小页面对齐

tag-icon • tag-icon vertical-alignment minipage
将一个小页面与另一个拉伸的小页面对齐

我对这个选项有点问题minipage。我想做的是将第一个minipage与第二个对齐，因为第二个是化学方程式。所以这是我的代码

\documentclass[a4paper,12pt]{scrartcl}
\usepackage[ngerman]{babel}
\usepackage{bpchem}
\usepackage[version=3]{mhchem}
\usepackage{chemexec} 
\begin{document}
\begin{minipage}[t]{.25\linewidth}
Oxidation:\\ 
Reduktion: \\
Redoxreaktion:\\
\end{minipage}%
\begin{minipage}[t]{.5\linewidth}
$\ce{ \ox{+II}{Mn} \op[2] + 4 \ox{-II}{O} \, \ox{+I}{H} \om[] -> \ox{+VI}{Mn} \, \ox{-II}>{O}_{2(s)} + 2 \ox{+I}{H}_2 \, \ox{-II}{O} + 2 e \om[] }$\\
$\ce{ \ox{+VII}{Mn} \, \ox{-II}{O}_4 \om[] + 2 \ox{+I}{H}_2 \, \ox{-II}{O}  + 3 e \om[] -> 3 \ox{+VI}{Mn} \, \ox{-II}{O}_{2(s)} + 4 OH \om[]}$ \\
$\ce{ \ox{+II}{Mn} \op[2] + 2 \ox{+VII}{Mn} \, \ox{-II}{O}_4 \om[] + 4 \ox{-II}{O} \, \ox{+I}{H} \om[] -> 5 \ox{+VI}{Mn} \, \ox{-II}{O}_{2(s)} + 2 \ox{+I}{H}_2 \, \ox{-II}{O}}$
\end{minipage}
\end{document}

答案1

你想要一个tabular：

\documentclass[a4paper,12pt]{scrartcl}
\usepackage[T1]{fontenc}
\usepackage[ngerman]{babel}
\usepackage[version=3]{mhchem}
\usepackage{chemexec}
\usepackage{bpchem}

\begin{document}
\begin{center}
\begin{tabular}{ll}
Oxidation: &
  \ce{ \ox{+II}{Mn} \op[2] + 4 \ox{-II}{O} \, \ox{+I}{H} \om[] ->
  \ox{+VI}{Mn} \, \ox{-II}>{O}_{2(s)} + 2 \ox{+I}{H}_2 \, \ox{-II}{O} + 2 e \om[] }
\\[1ex]
Reduktion: &
  \ce{ \ox{+VII}{Mn} \, \ox{-II}{O}_4 \om[] + 2 \ox{+I}{H}_2 \, \ox{-II}{O}  + 3 e \om[] ->
  3 \ox{+VI}{Mn} \, \ox{-II}{O}_{2(s)} + 4 OH \om[]}
\\[1ex]
Redoxreaktion: &
  \ce{ \ox{+II}{Mn} \op[2] + 2 \ox{+VII}{Mn} \, \ox{-II}{O}_4 \om[] + 4 \ox{-II}{O}
    \, \ox{+I}{H} \om[] ->
  5 \ox{+VI}{Mn} \, \ox{-II}{O}_{2(s)} + 2 \ox{+I}{H}_2 \, \ox{-II}{O}}
\end{tabular}
\end{center}
\end{document}

在此处输入图片描述

如果您希望论文中的所有反应都得到平等对待，请定义一个新环境：

\documentclass[a4paper,12pt]{scrartcl}
\usepackage[T1]{fontenc}
\usepackage[ngerman]{babel}
\usepackage[version=3]{mhchem}
\usepackage{chemexec} 
\usepackage{bpchem}
\usepackage{calc}

\newenvironment{reactionseries}
 {\par % start a new line
  \medskip % but leaving some space
  \noindent % flush left
  \begin{tabular}{
    @{} % no padding at the left
    p{.25\textwidth} % a quarter of the line for the description
    p{.75\textwidth-2\tabcolsep} % the rest for the reaction
    @{} % no padding to the left
  }% here the tabular starts
 }
 {\end{tabular}% here the tabular ends
  \par % new line
  \medskip % leave some space
 }

\begin{document}
\begin{reactionseries}
Oxidation: &
  \ce{ \ox{+II}{Mn} \op[2] + 4 \ox{-II}{O} \, \ox{+I}{H} \om[] -> 
  \ox{+VI}{Mn} \, \ox{-II}>{O}_{2(s)} + 2 \ox{+I}{H}_2 \, \ox{-II}{O} + 2 e \om[] }
\\[1ex]
Reduktion: &
  \ce{ \ox{+VII}{Mn} \, \ox{-II}{O}_4 \om[] + 2 \ox{+I}{H}_2 \, \ox{-II}{O}  + 3 e \om[] ->
  3 \ox{+VI}{Mn} \, \ox{-II}{O}_{2(s)} + 4 OH \om[]}
\\[1ex]
Redoxreaktion: &
  \ce{ \ox{+II}{Mn} \op[2] + 2 \ox{+VII}{Mn} \, \ox{-II}{O}_4 \om[] + 4 \ox{-II}{O} 
    \, \ox{+I}{H} \om[] ->
  5 \ox{+VI}{Mn} \, \ox{-II}{O}_{2(s)} + 2 \ox{+I}{H}_2 \, \ox{-II}{O}}
\end{reactionseries}
\end{document}

相关内容