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% Document Info
\author{yams-miny}
\pdfinfo{
  /Title (reactions-and-stoichiometry.pdf)
  /Creator (Cheatography)
  /Author (yams-miny)
  /Subject (Reactions and Stoichiometry Cheat Sheet)
}

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\noindent
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    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Reactions and Stoichiometry Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{yams-miny} via \textcolor{DarkBackground}{\uline{cheatography.com/124058/cs/23637/}}}
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\noindent
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}yams-miny \\
  \uline{cheatography.com/yams-miny} \\
  \end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 4th July, 2020.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{4}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Formulae}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Relative Isotopic Mass} \tn 
\mymulticolumn{1}{x{3.833cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}Mass of 1 atom of an isotope of an element / 1/12 the mass of 1 atom of carbon-12 isotope} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{Relative Atomic Mass} \tn 
\mymulticolumn{1}{x{3.833cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}Avg mass of 1 atom of an element / 1/12 the mass of 1 atom of carbon-12 isotope} \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Relative Molecular/Formula Mass} \tn 
\mymulticolumn{1}{x{3.833cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}Avg mass of 1 molecule/formula unit of a substance / 1/12 the mass of 1 atom of carbon-12 isotope} \tn 
% Row Count 10 (+ 4)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{Mr} \tn 
\mymulticolumn{1}{x{3.833cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}Sum of Ar of atoms in the molecular formula} \tn 
% Row Count 12 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Empirical Formula} \tn 
\mymulticolumn{1}{x{3.833cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}Simplest formula which shows ratio of atoms of different elements in the compound} \tn 
% Row Count 15 (+ 3)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{Molecular Formula} \tn 
\mymulticolumn{1}{x{3.833cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}Formula which shows actual number of atoms of each element in one molecule of the compound} \tn 
% Row Count 18 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Relative formula mass is used for ionic compounds \newline Relative masses have no units as they are ratios of 2 masses}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Calculations using Volume of Gases}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{Avogadro's Law: \newline % Row Count 1 (+ 1)
Equal volume of all gases, under the same temperature and pressure, contain the same number of particles (atoms or molecules) \newline % Row Count 4 (+ 3)
Gases in a balanced equation: Volume ratio = Mole ratio  \newline % Row Count 6 (+ 2)
Molar Volume, Vm: \newline % Row Count 7 (+ 1)
Volume occupied by 1 mole of the gas at a specific T\&P \newline % Row Count 9 (+ 2)
Standard T\&P : 273K (0 degree celsius), 1 bar (100 kPa), 22.7dm\textasciicircum{}3/mol \newline % Row Count 11 (+ 2)
Room T\&P: 293K (20 degree celsius), 1 atm (101 kPa), 24 dm\textasciicircum{}3/mol% Row Count 13 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{Volumes of gases are dependent on T\&P hence these conditions must be specified}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Stoichiometry}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{Stoichiometry: Quantitative aspects of chemical formulae \& reactions  \newline % Row Count 2 (+ 2)
Limiting reactants are completely consumed in the reaction and limit how much products can form. \newline % Row Count 4 (+ 2)
Percentage Yield =  \newline % Row Count 5 (+ 1)
Actual yield/mass or amount of product formed /  \newline % Row Count 6 (+ 1)
Theoretical yield/mass or amount of product formed  \newline % Row Count 8 (+ 2)
x 100\%% Row Count 9 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

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\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Types of Reactions}}  \tn
% Row 0
\SetRowColor{LightBackground}
Precipitation Reaction & Reactions which involve formation of insoluble solid (ppt) from reaction of 2 solutions \tn 
% Row Count 4 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
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\begin{tabularx}{3.833cm}{x{1.20155 cm} x{2.23145 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Types of Reactions}}  \tn
% Row 0
\SetRowColor{LightBackground}
Precipitation Reaction & Reactions which involve formation of insoluble solid (ppt) from reaction of 2 solutions, Separation by filtration or decanting \tn 
% Row Count 5 (+ 5)
% Row 1
\SetRowColor{white}
Thermal Decomposition & Chemical reaction caused by heat, Compounds break down into 2 or more substances \tn 
% Row Count 9 (+ 4)
% Row 2
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Acid-Base & Elaboration in a later segment \tn 
% Row Count 11 (+ 2)
% Row 3
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Redox & Elaboration in a later segment \tn 
% Row Count 13 (+ 2)
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\begin{tabularx}{3.833cm}{X}
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\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Calculations using Concentrations}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{When a solute is dissolved in a solvent, a solution is formed \newline % Row Count 2 (+ 2)
If the solvent is water, an aqueous solution is formed \newline % Row Count 4 (+ 2)
The concentration of a solution (mol dm\textasciicircum{}-3) shows the amt of solute dissolved in a given volume of solution  \newline % Row Count 7 (+ 3)
Standard solution: Solution whose concentration is accurately known \newline % Row Count 9 (+ 2)
{[}X{]} - Amt of X (mol) / V of solution (dm\textasciicircum{}3) \newline % Row Count 10 (+ 1)
No of moles of solute, n:  \newline % Row Count 11 (+ 1)
- {[}solute{]} (mol dm\textasciicircum{}-3) \textasciicircum{}X Volume (dm\textasciicircum{}3\textasciicircum{})  \newline % Row Count 12 (+ 1)
- Mass of X (g) / Molar mass of X (g/mol) \newline % Row Count 13 (+ 1)
When a solution is {\bf{diluted}} (by adding more solvent), the concentration of the solution decreases but no. of moles of solute in the diluted solution remains unchanged% Row Count 17 (+ 4)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{1dm\textasciicircum{}3\textasciicircum{} = 1000cm\textasciicircum{}3\textasciicircum{}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Acid-Base Reactions}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Arrhenius Theory of Acids \& Bases}} \newline % Row Count 1 (+ 1)
- An acid is a substance that dissociates in water to produce H3O+(aq) \newline % Row Count 3 (+ 2)
- A base is a substance that dissociates in water to produce OH- (aq) \newline % Row Count 5 (+ 2)
Neutralisation: H+ (aq) + OH-(aq) -\textgreater{} H2O (l) \newline % Row Count 6 (+ 1)
Limitation: Aqueous solutions only \newline % Row Count 7 (+ 1)
{\bf{Bronsted-Lowry Theory of Acids \& Bases}} \newline % Row Count 8 (+ 1)
- An acid is defined as any species which donates a proton, H+. It must thus contain Hi n its formula  \newline % Row Count 11 (+ 3)
- A base is defined as any species which accepts a proton, H+. It must contain a lone pair of electrons to bind the H+ ions  \newline % Row Count 14 (+ 3)
{\emph{Bronsted-Lowry acid-base reaction}} involves the transfer of a proton from an acid to a base. They do no occur only in aq solutions but also between gases and non-aq systems. \newline % Row Count 18 (+ 4)
{\emph{Limitation}}: Does not address why substances such as BF3 or AlCl3 do not contain any H atom but are known to behave as acids  \newline % Row Count 21 (+ 3)
 {\bf{Lewis Theory of Acid \& Bases}} \newline % Row Count 22 (+ 1)
- An acid is a species that accepts an electron pair, e.g. BF3 \newline % Row Count 24 (+ 2)
- A base is a species that donates an electron pair, e.g. NH3 \newline % Row Count 26 (+ 2)
{\emph{Lewis acid-base reaction}} can be viewed as a transfer of a pair of electrons from the base to the acid  \newline % Row Count 29 (+ 3)
Limitation: Too general \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
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\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Acid-Base Reactions (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{The 3 models can be used to interpret different acid-base systems \newline % Row Count 2 (+ 2)
BL and L theories - Describe specific acid-base reactions \newline % Row Count 4 (+ 2)
Arr theory - Whether isolated substances are acids, bases, or neither% Row Count 6 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{3.833cm}}{In aqueous solution, H+ does not exist on its own. It forms a dative bond with a water molecule to form H3O+, called hydronium or hydroxonium ion. Chemists often use H+ and H3O+ interchangeably o refer to the elevated H+ ion.  \newline  \newline Best to use Bronsted-Lowry theory wherever possible for an acid-base reaction, and apply Lewis theory only when reaction does not involve proton transfer}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{3.833cm}{x{1.33887 cm} x{2.09413 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Redox Reactions}}  \tn
% Row 0
\SetRowColor{LightBackground}
Redox Reaction & Reaction that involves reduction and oxidation simultaneously \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
Reduction & Process whereby a substance gains electrons, resulting in a decrease in OSN \tn 
% Row Count 7 (+ 4)
% Row 2
\SetRowColor{LightBackground}
Oxidation & Process whereby a substance loses electrons, resulting in an increase in OSN \tn 
% Row Count 11 (+ 4)
% Row 3
\SetRowColor{white}
Reducing agent (reductant) & Substance that gives electrons to another, itself being oxidised in the process \tn 
% Row Count 15 (+ 4)
% Row 4
\SetRowColor{LightBackground}
Oxidising agent (oxidant) & Substance that takes in electrons from another, itself being reduced in the process \tn 
% Row Count 19 (+ 4)
% Row 5
\SetRowColor{white}
\seqsplit{Disproportionation} & Redox reaction in which the same substance is both oxidised and reduced \tn 
% Row Count 22 (+ 3)
% Row 6
\SetRowColor{LightBackground}
Oxidation Number (OSN) & Number of electrons to be added or subtracted from an atom in a combined state to convert it to elemental form \tn 
% Row Count 27 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{3.833cm}}{Acronym: OIL RIG \newline When writing the OSN, +/- signs must be stated before the number}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
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\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Rules of Assigning OSN}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{3.833cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/yams-miny_1593862060_Oxidation+Numbers+Rules+for+Assigning+Oxidation+States.jpg}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Balancing Redox Reactions}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Method 1}} \newline % Row Count 1 (+ 1)
1. Balance elements that were oxidised or reduced \newline % Row Count 2 (+ 1)
2. Balance O with H2O \newline % Row Count 3 (+ 1)
3. Balance H with H+ \newline % Row Count 4 (+ 1)
4. Balance charges with electrons \newline % Row Count 5 (+ 1)
{\bf{Method 2}} \newline % Row Count 6 (+ 1)
1. Balance elements that were oxidised or reduced \newline % Row Count 7 (+ 1)
2. Add electrons (OSN x No of that element) \newline % Row Count 8 (+ 1)
3. Balance O with H2O \newline % Row Count 9 (+ 1)
4. Balance H with H+ \newline % Row Count 10 (+ 1)
{\bf{Method 3}} \newline % Row Count 11 (+ 1)
1. Balance elements that were oxidised or reduced \newline % Row Count 12 (+ 1)
2. Electrons gained = Electrons lost \newline % Row Count 13 (+ 1)
    - a x 7e = b x 2 x 1e  \newline % Row Count 14 (+ 1)
    - Same on LHS and RHS, figure out what a and b are, multiply the relevant coefficients \newline % Row Count 16 (+ 2)
3. Balance O with H2O \newline % Row Count 17 (+ 1)
4. Balance H with H+% Row Count 18 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{3.833cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Balancing Redox Reactions}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{{\bf{Method 1}} \newline % Row Count 1 (+ 1)
1. Balance elements that were oxidised or reduced \newline % Row Count 2 (+ 1)
2. Balance O with H2O \newline % Row Count 3 (+ 1)
3. Balance H with H+ \newline % Row Count 4 (+ 1)
4. Balance charges with electrons \newline % Row Count 5 (+ 1)
{\bf{Method 2}} \newline % Row Count 6 (+ 1)
1. Balance elements that were oxidised or reduced \newline % Row Count 7 (+ 1)
2. Add electrons (OSN x No of that element) \newline % Row Count 8 (+ 1)
3. Balance O with H2O \newline % Row Count 9 (+ 1)
4. Balance H with H+ \newline % Row Count 10 (+ 1)
{\bf{Method 3}} \newline % Row Count 11 (+ 1)
1. Balance elements that were oxidised or reduced \newline % Row Count 12 (+ 1)
2. Electrons gained = Electrons lost \newline % Row Count 13 (+ 1)
    - a x 7e = b x 2 x 1e  \newline % Row Count 14 (+ 1)
    - Same on LHS and RHS, figure out what a and b are, multiply the relevant coefficients \newline % Row Count 16 (+ 2)
3. Balance O with H2O \newline % Row Count 17 (+ 1)
4. Balance H with H+% Row Count 18 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}