Harvesting CaCo2 in a precipitate reaction results in less grams desired amount (2 grams)
Chemistry 1
February 3, 2014
Harvesting CaCo2 in a precipitate reaction results in less grams desired amount (2 grams)
Abstract:
The objective of this experiment was to synthesize 2.00 grams of CaCo2 for a hypothetical manufacturing company using the two least detrimental (in regards to health and the environment) and least expensive chemicals possible through a precipitate reaction. A precipitate reaction happens when cations and anions of certain aqueous solutions react and combine forming a precipitate, which is an insoluble ionic solid. In this precipitate reaction we dissolved both chosen chemicals (powder Calcium Chloride dihydrate and powder Sodium Carbonate) in distilled water and then mixed them together in a single beaker. Finally we filtered this mixture and let all the water evaporate leaving behind solely the precipitate. In order to determine the amounts needed of each reactant we had to use stoichiometry. First we had to set up a balanced equation of the precipitate reaction that yielded 2.00 grams of CaCo3, then we looked at the ratio of moles reactant to moles product. We then substituted the ratio of the molar mass of each reactant to molar mass of the product in for each mole. We divided this to find out how many grams of reactant for every one gram of product, and finally multiplied this by two. Our final results were lower than the desired 2.00 grams.
Balanced Equation:
CaCl2 X 2H2O + Na2CO3 x H2O → CaCO3 + 2NaCl + 3H2O Background Information:
We chose Calcium Chloride dihydrate because it is soluble in water, not very expensive and only mildly toxic (LD50 1000mg/kg). In addition we chose Sodium Carbonate because it is soluble in water, inexpensive and only less toxic (LD50 4090mg/kg). Our Product (CaCo3) is slightly soluble in water; in acids with evolution of carbon dioxide, it is a white color and has no odor with a toxicity of LD50 6450mg/kg.
Calculations:
1 mol CaCl2 X 2H2O/ 1 mol CaCo3 X 147.01056 g CaCl2 X 2H2O/
February 3, 2014
Harvesting CaCo2 in a precipitate reaction results in less grams desired amount (2 grams)
Abstract:
The objective of this experiment was to synthesize 2.00 grams of CaCo2 for a hypothetical manufacturing company using the two least detrimental (in regards to health and the environment) and least expensive chemicals possible through a precipitate reaction. A precipitate reaction happens when cations and anions of certain aqueous solutions react and combine forming a precipitate, which is an insoluble ionic solid. In this precipitate reaction we dissolved both chosen chemicals (powder Calcium Chloride dihydrate and powder Sodium Carbonate) in distilled water and then mixed them together in a single beaker. Finally we filtered this mixture and let all the water evaporate leaving behind solely the precipitate. In order to determine the amounts needed of each reactant we had to use stoichiometry. First we had to set up a balanced equation of the precipitate reaction that yielded 2.00 grams of CaCo3, then we looked at the ratio of moles reactant to moles product. We then substituted the ratio of the molar mass of each reactant to molar mass of the product in for each mole. We divided this to find out how many grams of reactant for every one gram of product, and finally multiplied this by two. Our final results were lower than the desired 2.00 grams.
Balanced Equation:
CaCl2 X 2H2O + Na2CO3 x H2O → CaCO3 + 2NaCl + 3H2O Background Information:
We chose Calcium Chloride dihydrate because it is soluble in water, not very expensive and only mildly toxic (LD50 1000mg/kg). In addition we chose Sodium Carbonate because it is soluble in water, inexpensive and only less toxic (LD50 4090mg/kg). Our Product (CaCo3) is slightly soluble in water; in acids with evolution of carbon dioxide, it is a white color and has no odor with a toxicity of LD50 6450mg/kg.
Calculations:
1 mol CaCl2 X 2H2O/ 1 mol CaCo3 X 147.01056 g CaCl2 X 2H2O/