The chemicals on the tail end are called the reactants and the chemicals on the other side are called products. An example of this given by Coefficients (2008) is 2H2 + O2 > 2H2O. In this example, 2H2 + 02 is the reactants and 2H2O is the product. Also, > is the sign for yield. The big 2s in front of H2 and H2O are called coefficients. In this case, the first 2 indicates that there are 2 molecules of H2, which also means that there are 4 atoms of hydrogen in the reactant part of the equation. The other 2 signifies that there are 2 molecules of H2O as the product.
This means that every molecule of H2O that contains 2 atoms of hydrogen contains 4 hydrogen atoms and 2 oxygen atoms. According to Chemistry Formulas (2005), the subscripts are used to signify the number of atoms of an element in a compound. In this example, O2 means that there are 2 atoms of oxygen. To balance an equation, How can (n. d. ), says that an element inventory is first made. For example, if there was the equation H2 + O2 > H2O, it would be written down that on the left side there are 2 hydrogen and 2 oxygen, and on the right side there are 2 hydrogen and 1 oxygen.
This is determined by the subscripts. Next, the guess and check method is used to determine the correct coefficients. When doing this, the inventory is updated. A common chemical reaction is a precipitation reaction. As explained by (McGraw-Hill. (n. d. ), this reaction involves two ore more solutions combining and resulting in an insoluble product. This reaction happens because of strong attractive forces that certain ions have for each other, which combine and fall out of the solution in the form of a solid. When two aqueous solutions combine and form a solid, precipitate is formed.
As explained by Zahm (2008), a mole ratio is used to find an amount of a chemical from a known amount of another chemical. For example, if someone needed to know how much nitrogen is required to react with 9 moles of hydrogen, a ratio would be set up that looks like this: N2 : 1 = x H2 3 9 When x is solved, it shows that 3 moles of nitrogen are required to react with 9 moles of hydrogen. N2 represents the unknown, H2 represents the known, 1 represents the amount of the unknown, and 3 represents the amount of the known.
An aqueous solution is a solution where water us the dissolving solvent. In an equation, it is usually represented by (aq) , as told by Aqueous solution. (2005). Examples of aqueous solutions are saltwater and soda. According to Stochiometry (n. d. ), a limiting reactant is the reactant in a chemical that limits the amount of what can be formed. When all of the limiting reactant is consumed, the reaction will stop. There are six types of reactions. They are explained by The Six (n. d. ). The first is combustion.
This is when oxygen combines with another compound, forming water and carbon dioxide. The second is synthesis, where two or more simple compounds mix, and form a complicated one. The third is decomposition, where a complex molecule is broken down to make simpler ones. The fourth is single displacement, where one element switches places with an element in the same compound. Double displacement is when the cations and anions in different molecules trade places. The sixth is acid-base, which is a special type of double displacement. It occurs when a base and an acid react with each other.
The main objective of this lab is to find the correct ratio of reactants in the reaction of Pb(NO3)2(aq) with K2Cr2O7 to form PbCr2O7(s) + KNO3(aq). Another objective of the lab is to learn how to balance equations. The correct mole ratio of the compounds is one to one because when the equation is correctly balanced out, it is Pb(NO3)2(aq) + K2Cr2O7(aq) > PbCr2O7(s) + 2KNO3(aq). The well ratio of reactants that should result in the most product will be well 5 because the balanced equation has the ratio one to one, and well 5 has 5 drops of lead nitrate and 5 drops of potassium dichromate, which simplifies to 1 to 1.
Methods 1) Obtain about 10 mL of K2Cr2O7(aq),and about 10mL of Pb(NO3)2(aq). (These amounts do not have to be exact, but should be close so as not to waste any of these solutions later. ) 2) Using micropipettes as droppers, mix the two reagents by mixing drops according to the table #1 below. Mix the two reactants in the wells of the microchemistry plates. Be certain to use the line of wells near the edge of the plate such that you can see the results easily. NOTE the wells are numbered on the plate and should correspond to the data table.
3) For each well, mix WELL with the microspatula after adding the reagents. (Remember to clean off spatula after each stirring so as not to mix extra reactants not accounted for by well#) After all wells have been mixed, allow the plate to settle for about five minutes and observe relative heights of the solid product, PbCr2O7(s). Indicate which well had the most precipitate. This should be done by ranking with a 1 for the well with the most precipitate product and a 9 for the one with the last precipitate product.
If your results are inconclusive, repeat the procedure. 4) Observe the degree of color intensity of each well. A scale should also be used for this with 1 corresponding to the lightest and 9 corresponding to the darkest. Record your observations of the color in addition to the relative amount of precipitate formed. 5) Repeat the entire experiment after thoroughly cleaning out your microchemistry plate! Results Table #1: Experimental Groups of Reactant Concentrations Well # 1 2 3 4 5 6 7 8 9.
Drops of Pb(NO3)2(aq) 1 2 3 4 56 7 8 9 Drops of K2Cr2O7(aq) 9 8 7 6 5 4 3 2 1 Ranked intensity 1 3 7 8 9 6 5 4 2 Ranked precipitate 1 3 7 8 9 6 5 4 2 This table shows the number of drops of lead nitrate and potassium dichromate in 9 different wells. It also shows the intensity and amount of precipitate of what was formed on a scale of 1-9, 9 being the most precipitate and the darkest intensity, and 1 being the last precipitate and the lightest intensity. Well 5 had the most precipitate and the darkest intensity, and well 1 was the opposite. Discussion.
The main objective of this lab was to determine the correct ratio of reactants in the reaction of Pb(NO3)2(aq) with K2Cr2O7 to form PbCr2O7(s) + KNO3(aq). This was successfully done, both theoretically and experimentally. When the equation was balanced, it showed that there was a one to one ratio of reactants. This was the theoretical ratio. In the experiment, well 5 had the most precipitate, which meant it had very little or nothing that wasnt reacted. In well 5, there was five drops of both lead nitrate and potassium dichromate.
This meant that this well had a 5 to 5 ratio, which simplified to a one to one ratio. Therefore, the hypothesis was correct. The amount of precipitate in the wells was rated on a scale of 1 to 9, 1 being the least and 9 being the most. With evidence in Table #1, wells 1 and 2 had the least amount of precipitate. This was because they had the ratios that were the most incorrect, since the correct one was 5 to 5, and their ratios were 1 to 9 and 9 to 1. The ratio of the compounds in well 4 was the closest to the correct ratio, resulting in the second most amount of precipitate.
In general, the father the well was from well 5, the less precipitate there was for that ratio. Each well had a certain amount of drops of lead nitrate and potassium dichromate, which also meant a certain ratio of the two. These ratios are representative of possible ratios in balancing the chemical equation. Although the hypothesis was correct, there could have been a few possible sources of error. One error source could have been that the compounds were not completely dissolved in water. This would cause the reactions to not happen as strongly, resulting in wrong results.
Another possibility is that the droppers could have been contaminated, which would also result in wrong results. Also, if the two compounds werent mixed with a toothpick, the compounds wouldnt combine enough. In addition to the main objective being met, the other objective, which was to balance equations, was also met. There is evidence of this in the hypothesis, where the chemical equation used in the lab is correctly balanced. The hypothesis turned out to be correct, and both objectives were met. In conclusion, this was a successful lab.
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