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When calcium carbonate is heated strongly carbon dioxide gas is released CaCO3CaO CO2 what is the volume of CO2 measured stpis produced if 1 grams of caco3 is heated?
When 1 gram of calcium carbonate (CaCO3) is heated, it decomposes to produce calcium oxide (CaO) and carbon dioxide (CO2). The molar mass of CaCO3 is approximately 100 g/mol, so 1 gram corresponds to about 0.01 moles of CaCO3. According to the reaction, 1 mole of CaCO3 produces 1 mole of CO2; thus, 0.01 moles of CaCO3 will produce 0.01 moles of CO2. At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters, so the volume of CO2 produced is approximately 0.01 moles × 22.4 L/mol = 0.224 liters, or 224 mL.
What are the reactants and products in this equation CaCO3CaO CO2?
Its hard to tell from the way you wrote it, but if its supposed to be CaCo3---> CaO + CO2, then CaCo3 is the reactant and CaO and CO2 are the products. Its simply that the elements on the left of the arrow are reactants and the ones on the right are products.
When calcium carbonate is heated strongly carbon dioxide gas is released CaCO3CaO CO2 what is the volume of CO2 measured stpis produced if 15.2 grams of caco3 is heated?
The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CO2 is 44.01 g/mol. By using stoichiometry, you can calculate that 15.2 grams of CaCO3 would produce 6.51 grams of CO2. Using the ideal gas law, you can then convert the mass of CO2 to volume using its molar volume at STP (22.4 L/mol). The volume of CO2 produced would be around 3.32 liters.
