1. The first step of my calculations was finding the number of moles of CaCl2 and NaOH added in each test. The volume of CaCl2 is an increasing number with a concentration of 1.0M. The volume of NaOH is constant for all four tests, but the concentration is 2.4588M. To find the number of moles of each reactant added, volume in liters was multiplied by the molarity (concentration).
2. The second step is about finding the theoretical yield, which will help to determine the correct amount of Ca(OH)2 can be made in chemical reaction. However, before doing this, it’s necessary to find whether CaCl2 or NaOH is a limiting reagent. For each test, the limiting reagent is found by multiplying the number of moles of the reactant by 1 mole of Ca(OH)2 and dividing then by a number of moles of reactant from the reaction. The lowest answer in each test will be the limiting reagent. To find a theoretical yield, the limiting reagent was multiplied by the molar mass of Ca(OH)2 and
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The third step is identifying the limiting reagents for each test. It can easily be classified by the smallest number. The higher number will be the excess.
4. After identifying reagents and finding the theoretic yield, it’s possible to find the excess reagent mass and number of moles for each test. To do this, the smallest mole number of Ca(OH)2 was subtracted from the highest one. The result is used to find the amount of moles excess, by multiplying it to the corresponding number of moles of excess reagent and dividing then by 1 mole of Ca(OH)2. After finding the answer in moles, it’s possible to find the number of grams by following the rules of conversion factors from moles to grams.
5. The fifth step was graphing the volume of CaCl2 with respect to the mass of dried Ca(OH)2. The x-axis shows the volume of CaCl2 and y-axis shows the mass of dried Ca(OH)2. The plateau region is supposed to start from the volume point of 5mL, but due to the mistakes in the lab experiment it’s decreasing