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diff --git a/cer/sem1/yield/limiting-reactant.log b/cer/sem1/yield/limiting-reactant.log
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diff --git a/cer/sem1/yield/limiting-reactant.pdf b/cer/sem1/yield/limiting-reactant.pdf
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diff --git a/cer/sem1/yield/limiting-reactant.tex b/cer/sem1/yield/limiting-reactant.tex
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+\input ../cer.tex
+
+\name{Holden Rohrer}
+\course{FVS Chemistry AB 19.3}
+\teacher{Kerr}
+
+\question{While observing a chemical reaction, how can you tell which reactant is limiting?}
+\claim{If a substance is the limiting reactant, then it will be fully consumed by the time the reaction completes because it is the reactant that reacts completely and the reaction cannot proceed further.}
+
+\long\def\makedata{
+  \vskip 0pt \noindent{\big Trial 1}
+  
+  \def\width{0.5in}
+  \def\text##1{\hbox{\vbox{\hsize\width \tolerance=10000 \hbadness=10000 \noindent##1}}}
+  \def\style{\vrule width 1pt \head \hfil \text{####} \hfil & \vrule width 0.05pt \hfil #### \hfil & \vrule width 0.05pt \hfil #### \hfil & \vrule width 2pt \hfil #### \hfil & \vrule width 0.05pt \hfil #### \hfil \vrule width 1pt \crcr}
+  \def\cr{\crcr \noalign{\hrule height 0.05pt}}
+  \halign{ \span\style
+    \noalign{\dimen1=\hsize \advance \dimen1 by -0.65in \hfil \vbox{\moveright0.185in\vbox{\hrule width \dimen1 height 1pt}} \vskip -5pt}
+    \omit & \omit \vrule width 0.05pt \ $3CuCl_2(aq)$ & \omit + $2Al(s)$ & \omit $\rightarrow$ $3Cu(s)$ & \omit + $2AlCl_3(aq)$ \vrule width 1pt \cr
+    Measur\-ed Mass (g) & 2.50 & 0.50 & N/A & N/A \cr
+    Molar Mass (g/mol) & 134.45 & 26.98 & N/A & N/A \cr
+    Actual Moles (mol) & 0.019 & 0.019 & N/A & N/A \cr
+  }
+
+  \smallskip The limiting reactant is copper chloride because there was still aluminum left over after the reaction stopped. The reaction has a molar ratio of $CuCl_2:Al=3:2$, which is confirmed by this trial (copper chloride is limiting because $CuCl_2:Al$ ratio $= 1:1 < 3:2$)
+
+  \bigskip \noindent{\big Trial 2}
+
+  \halign{ \span\style
+    \noalign{\dimen1=\hsize \advance \dimen1 by -0.59in \hfil \vbox{\moveright0.255in\vbox{\hrule width \dimen1 height 1pt}} \vskip -5pt}
+    \omit & \omit \vrule width 0.05pt \ $3CuCl_2(aq)$ & \omit + $2Al(s)$ & \omit $\rightarrow$ $3Cu(s)$ & \omit + $2AlCl_3(aq)$ \vrule width 1pt \cr
+    Measur\-ed Mass (g) & 2.50 & 0.25 & \vbox{\tiny \parindent=0in \hsize 0.57in \raggedright Filter paper alone: 0.27g \par Filter Paper + Cu: 0.98g \par Cu alone: 0.71g} & N/A \cr
+    Molar Mass (g/mol) & 134.45 & 26.98 & 63.55 & N/A \cr
+    Actual Moles (mol) & 0.019 & 0.0093 & 0.011g Cu & N/A \cr
+  }
+
+  \smallskip The limiting reactant is aluminum because all of the aluminum dissolved, and the molar ratio of $CuCl_2$ to $Al > 3:2$.
+}
+
+\evidence{
+This evidence shows that the claim is true because whe\-re copper chloride should be the limiting reactant theoretically, it completely reacted and the aluminum was left over. Similarly, trial 2 (limiting reactant should theoretically be aluminum because of the ratio) saw that the aluminum fully dissolved. When comparing the two trials, it can be clearly seen that the "removal" of aluminum caused the substance which didn't allow the reaction to be completed to change. 
+}
+\justification{
+This result makes sense because the limiting reactant would allow the reaction to continue if more were added, by definition. This can lead us to conclude that no more of the limiting reactant is available at the end of the reaction because then adding more of the limiting reactant could not possibly allow the reaction to continue. The reactant being unavailable at the end of the reaction is equivalent to the limiting reactant being fully consumed by reacting completely in all of its particles.
+}
+
+\makeheader
+\makedoc
+\bye
diff --git a/cer/sem1/yield/texput.log b/cer/sem1/yield/texput.log
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