This week we applied net ionic equations to a more complicated equation which involved dealing with charges and balancing electrons. Redox equations are straightforward yet they still seem to give me trouble on the little things. I will now try to stay on top of all the equations as we apply acids and bases in upcoming sections.
Chemical reactions can be detected fairly easily in chemistry. Three indicators of a chemical reaction are gas production, color change, and temperature change. In the mini lab we did today we set up a series of reactions each with separate indicators. In section one there was two reactions. Reaction one was when steel wool combines with oxygen. The balanced equation for that reaction is as shown 4Fe=3O2 —> Fe2O3. The second reaction in this section was with calcium oxide and water. My groups observations for this reaction was that it turned a milky white. When we added the Bromothymol blue indicator the solution turned blue indicating the solution was a base. The balanced equation is CaO+H2O—>Ca(OH)2. The reactions in this section were similar because they started with 2 reactants and then ended with just one. Section two consisted of two reactions also. Reaction 3 was the decomposition of hydrogen peroxide. The balanced equation was 2H2O2 (MnO2- catalyst)—> 2H2O+O2. Reaction four was decomposition of sodium hydrogen bicarbonate. The balanced equation is 2NaHCO3—> H2O=NaCO3+CO2. These two reactions started with one reactant and ended with two. Section three had two reactants as well. Reaction five was calcium and water. This reaction turned white and heated up. The BTB indicated it was a base because it stayed blue. The equation is CA+H2O—>H2+CA(OH)2. Reaction six was zinc and lead (II) nitrate: Zn+Pb(NO3)2—>Pb+Zn(NO3)2. These two reactions are similar because they went from element to compound both times. Section four had 3 parts. Reaction seven was sodium carbonate and barium nitrate: Na2(O3+Ba(NO3)2—>NaNO3+BaO3. Reaction eight was Lead (II) Nitrate and Potassium Iodide: Pb(NO3)2+2KI—>2KNO3+PbI2. Reaction nine: CACO3+@HCL—>CO2+H2O+CaCl2. These reactions all switch their reactants and precipitants around and that’s what they have in common. Section 5 has two reactions. Reaction ten: CH4+2O2—>CO2+2H2O. Reaction eleven: 4C2H5O2+9O2—>8CO2+10H2O. Section five are all combustion meaning their precipitants are always CO2 and H2O. You can tell these reactions are chemical reactions because each of them has at least one of the three indicators. Each section has a unique property, or set of reactants, precipitants, etc. Knowing this you can create a unique model for each reaction because it is predictable. Reactions can be predicted because they have set compounds or elements in the balanced equation. This means that you can predict the precipitant because you have to be able to balance your equation and use those compounds, so yes it is predictable.
Sources: Class experiment
This week in chemistry we talked about stoichiometry. I learned that you can determine the amounts of compounds that can be derived from a certain chemical equation using stoichiometry. We also took a test this week. From that I learned that I didn’t quite grasp the hint but I understand my mistakes. Next I am going to study more and ask questions to make sure I fully grasp each section.
In this explore lab we experimented with the reactions of white powders and vinegar. We had set amounts of one variably and the other variable amount varied. We found the class data to be unpredictable because some people had errors throughout the lab, but what we could decipher from the data was that the graph of the data was linear at first and then eventually plateaued. The reason for this was, for example, if you add baking soda you add to a constant amount of vinegar it will keep increasing but eventually there will be too much baking soda to to little vinegar. This is where the limiting reactant comes into play. In this example the vinegar is the limiting reactant. This can be predictable for set elements and solutions because you have data that you can infer the future possibilities with. The limiting reactant is how much of a substance you need to make an experiment possible. The excess reactant is how much you have left over from the other compound or solution. These reactants can always be determined by the balanced equation. You can calculate how much of the excess reactant is left over by using the molar ratio in the balanced equation.
For more information on Stoichiometry Check out this video!
In this mini lab our chemistry class experimented with different white powders and liquids that when mixed together produced a gas. First it was vinegar and sodium bicarbonate. The result was fizzing which produced a gas. We had 1.405g of sodium bicarbonate, which when added to the vinegar produced 1.587g of CO2. We also had a percent error due to the fact that the solution fizzed over the top onto the table which caused a larger amount of gas to be recorded because that mass was lost from the mixture. We did this similar experiment three different times with 3 combinations added together. The result always being a 1:1 molar ratio. This means for every 1 mole of powder used, there should be one mole of gas produced. With the exception of the last experiment, all the experiments had this quality. For the last equation we had to balance the compounds. We put 2’s in front of NaHCO3 and CO2. This resulted in a 2:2 molar ratio because we had to balance the equations. Now this ratio does not apply to all chemical equations. Each different combination of compounds will have a different reaction causing more or less gas. But for that equation the theoretical ratio should be constant. There is a molar relationship between the two substances. Although it may change, there is always a line of best fit that tags along with the data. The theoretical relationship you should calculate with every experiment demonstrates the precise ratio one should get when doing the experiment.