1400 grams
3.5 moles CaCO3 (1 mole carbon/1 mole CaCO3) = 3.5 moles
The formula given shows that each formula unit or mole contains one calcium atom; therefore, 2.5 moles of calcium chloride contains 2.5 moles of calcium atoms.
Assuming each Tums tablet contains 500 mg of calcium carbonate, there is a total of 6 grams of calcium carbonate in one roll of Tums (12 tablets x 500 mg). To calculate the number of moles, divide the mass by the molar mass of calcium carbonate (100.09 g/mol), yielding around 0.06 moles of calcium carbonate in one roll of Tums.
This question is solved with the help of mole concept . 1 mole of anhydrous calcium carbonate weighs 40+12+48=100 gm . 1.25 mole of similar anhydrous calcium carbonate will be 100* 1.25 = 125 gm
1 mole CaCO3 (3 mole O/1 mole CaCO3) = 3 moles oxygen
1 mole of calcium carbonate produces 1 mole of carbon dioxide when it decomposes. Therefore, if 2.5 moles of calcium carbonate is consumed, 2.5 moles of carbon dioxide will be produced.
2,8 moles of calcium carbonate have 240,208 g.
3.5 moles CaCO3 (1 mole carbon/1 mole CaCO3) = 3.5 moles
To find the amount of calcium chloride produced, first calculate the moles of calcium carbonate and hydrochloric acid using their molar masses. Then, determine the limiting reactant and use stoichiometry to find the moles of calcium chloride produced. Finally, convert moles of calcium chloride to grams using its molar mass.
To calculate the mass of calcium oxide that can be produced from 4.7 kg of calcium carbonate, you need the molar masses of both calcium carbonate (CaCO₃) and calcium oxide (CaO). Additionally, you need to know the balanced chemical equation for the decomposition of calcium carbonate, which is: CaCO₃ → CaO + CO₂. Using this information, you can determine the moles of calcium carbonate and then use stoichiometry to find the corresponding mass of calcium oxide produced.
The number of moles of calcium carbonate are 3.5 moles. , there are 1 mole of calcium (Ca) atom, 1 mole of carbon (C) atom and 3 moles of oxygen (O) atoms.
To calculate the mass of calcium oxide produced from the decomposition of calcium carbonate, you need the molar masses of calcium carbonate (CaCO₃) and calcium oxide (CaO). Additionally, you must know the amount (in grams or moles) of calcium carbonate you are starting with. Using stoichiometry from the balanced chemical equation, you can determine the mass of calcium oxide that can be produced.
The formula given shows that each formula unit or mole contains one calcium atom; therefore, 2.5 moles of calcium chloride contains 2.5 moles of calcium atoms.
To calculate the grams of calcium carbonate needed, you first need to determine the moles of carbon dioxide produced (using the ideal gas law). Then, since 1 mole of CO2 is produced for every mole of CaCO3 consumed, you can convert moles of CO2 to moles of CaCO3. Finally, use the molar mass of CaCO3 to convert moles to grams.
Assuming each Tums tablet contains 500 mg of calcium carbonate, there is a total of 6 grams of calcium carbonate in one roll of Tums (12 tablets x 500 mg). To calculate the number of moles, divide the mass by the molar mass of calcium carbonate (100.09 g/mol), yielding around 0.06 moles of calcium carbonate in one roll of Tums.
Calcium carbonate, CaCO3 has formula mass of 40.1+12.0+3(16.0) = 100.1Amount of CaCO3 = 1.719/100.1 = 0.0172molThere are 0.0172 moles of calcium carbonate in a 1.719 gram pure sample.
The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol. Therefore, the mass of 0.5 moles of calcium carbonate would be 0.5 mol x 100.09 g/mol = 50.045 grams.