Before the experimenting phase of the lab, the students must first calculate the molarity of the given solutions so that exact reacting ratio will occur between the two solutions when mixed. One reactant will be constant, as in each lab the molarity and the volume of that reactant will remain the same. The other reactant, however, will vary. The volume of the reactant will change during every reaction for it's the independent variable in the lab and the objective of this lab is to study how the change in molarity affects the amount of product that will form. The product is the substance that forms from the reaction of the two reactants. Some information is given, such as the molarity and volume of the constant reactants and the volume of the varying reactant. The problem is to calculate the unknown molarity of the varying reactant. But before any calculations should be done, the first concept that needs to be understood is ‘exact reacting ratio'. Exact reacting ratio is when two solutions are combined, the ratio of moles are perfect so that both reactants will completely react. But if the mole ratio isn't perfect and the number of moles of each reactant has isn't proportional, then during the reaction, one of the chemicals will get used up before the other one.
[...] Before the experiment, the students must make their respective molar solutions. But first, they need to figure out how much of Na2SO4 is required to make their molar solutions, and they calculate this by using the following formula: 100.ml solution X 1 Liter sol. X amount of mole(molarity) X 322.25 g 1000.ml sol 1 Liter 1 mole Na2So4+10H2O In this experiment, the students are also trying to predict the amount of precipitate, or the substance that will fall out of solution during a reaction, that will form after the reaction. [...]
[...] The experimental phase of the lab starts out requiring the students to make molar solutions of their respective molarity of Na2SO4+10H2O solution. After finding the amount of Na2SO4+10H2O necessary for their molar solutions (see calculations), the student will then make their molar solution. In my case, I had to make a 100ml 0.0200 M solution of Na2SO4+10H2O. I calculated the amount of Na2SO4 needed for 0.0200 moles of Na2SO4+10H2O in 100.ml solution was 0.644 grams (see calculations), I massed the Na2SO4+10H2O to that amount, and poured it into a 100ml volumetric flask. [...]
[...] molar solution of Na2SO4+10 H20 for the experiment, make sure one of the molarity results in the exact reacting ratio. At least 3 must result in the assigned chemical being the limiting reactant. And At least 3 must result in the assigned chemical being the excess reactant. Set up a data table or the experiment and create a procedure before starting. Each student must then calculate the necessary mass of the solute needed for his or her own molar solution. [...]
[...] When my assigned 100.ml molar solution of Na2SO4+10H2O was mixed with the 10.ml solution of SrCl2+6H2O, it started to be transparent at first, but after a little swirling in the Erlenmeyer flask, the precipitate (which was SrSO4) started to solidify after around 20 seconds. It turned into a heterogeneous solution as the precipitate was suspended in the solution. The precipitate was milky, white, and cloudy. When dried, the precipitate was white and powdery. Experimental Errors Different quadruple beam balances were being used. [...]
[...] The filter paper with the dried precipitate will then be ready to be massed. Subtract the mass of the filter paper and the mass of the precipitate would be calculated. Then we pooled our data of precipitate yielded in our individual experiments to come up with a data table and then plot the experimental data as points on the limiting and excess reactant graph. A best-fit line is drawn and it allowed the students to compare and contrast the differences between theoretical data and the actual experimental data. [...]
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