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CaCO3, or calcium carbonate, has a molar mass of 100.09 grams per mole. This means there are roughly 566 grams of CaCO3.
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Calculate the number of moles in 27.50 grams CaCO3?
Approximally 0.275 moles. The molar mass of Ca C O3 is ~ 40+12+3*16 = 100g/mol 27.50g = x mol *100g/mol 27.50g/(100g/mol) = x mol 0.275 g/(g/mol) = x mol 0.275 mol = x mol
How many moles of oxygen atoms are in 4.25 moles of calcium carbonate the chief constituent of seashells?
Calcium Carbonate (CaCO3)I mol CaCO3 contains 3 mol Oxygen atomsso 4.25 mol CaCO3 will have 12.75 mol Oxygen Atoms.
What volume of a 0.10 mol dm-3 solution of HCl is required to react with 10.0 g of CaCO3?
2 HCl + CaCO3 = CaCl2 + CO2 + H2O Therefore it takes twice as many moles of HCl to react with CaCO3 Mol CaCO3 = mass / formulae weight = 10 / 40.08 + 14.01 + 3(16.00) = 10 / 102.09 = 0.0980 mol Mol HCl = 2 x 0.0980 = 0.1959 mol Volume HCl = mol / c = 0.1959 / 0.10 = 1.9591 L Volume required to react of HCl = 1.9591 L
How many grams of calcium carbonate are needed to produce 95.0 L of carbon dioxide?
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
How do you determine the number of kilograms of CaCO3 needed to neutralize 500 L of 1.40 M HCl?
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l) 500 L * (1.40 mol HCl/1 L HCl) * (1 mol CaCO3/2 mol HCl) * (100.09 g CaCO3/1 mol CaCO3) = 35031.5 g CaCO3 35031.5 g * (1 kg/1000 g) = 35.0315 kg Therefore, about 35.03 kilograms of calcium carbonate is needed to neutralize 500 liters of 1.40 M hydrochloric acid.
Calculate the number of moles in 27.50 grams CaCO3?
Approximally 0.275 moles. The molar mass of Ca C O3 is ~ 40+12+3*16 = 100g/mol 27.50g = x mol *100g/mol 27.50g/(100g/mol) = x mol 0.275 g/(g/mol) = x mol 0.275 mol = x mol
Calculate the mass in grams of 2.5 moles in calcium carbonate?
For this you need the atomic (molecular) mass of CaCO3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CaCO3= 100.1 grams2.50 moles CaCO3 × (100.1 grams) = 250.25 grams CaCO3
How many moles of oxygen atoms are in 4.25 moles of calcium carbonate the chief constituent of seashells?
Calcium Carbonate (CaCO3)I mol CaCO3 contains 3 mol Oxygen atomsso 4.25 mol CaCO3 will have 12.75 mol Oxygen Atoms.
What is the mass of CaCO3?
Molar mass of CaCO3 = 100.0869 g/mol
What volume of a 0.10 mol dm-3 solution of HCl is required to react with 10.0 g of CaCO3?
2 HCl + CaCO3 = CaCl2 + CO2 + H2O Therefore it takes twice as many moles of HCl to react with CaCO3 Mol CaCO3 = mass / formulae weight = 10 / 40.08 + 14.01 + 3(16.00) = 10 / 102.09 = 0.0980 mol Mol HCl = 2 x 0.0980 = 0.1959 mol Volume HCl = mol / c = 0.1959 / 0.10 = 1.9591 L Volume required to react of HCl = 1.9591 L
How many grams of calcium carbonate are needed to produce 95.0 L of carbon dioxide?
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
How many grams are in strontium chloride?
SrCl2 Strontium (Sr) = 87.62 grams/mol 2 Chlorine (Cl) = 70.9 grams/mol ----------------------------------------------------add Strontium chloride = 158.52 grams/mol ============================
What is the mass in grams of 0.250 mol of the common antacid calcium carbonate?
molar mass CaCO3= 100g/m 100 x 0.250= 25 g
What mass of calcium oxide will be obtained by heating 3 mol of caCO3?
3 mol CaCO3 will produce 3 mol CaO (and 3 mol CO2) on dry-heating. So 3 times the molar mass of CaO (56,1 g/mol) which is 168 g CaO
How many grams in 4.23 m mol of HBr?
The answer is 0,3422 grams.
How many grams of co2 are in 2l mol of the compound?
92.4 grams
How many grams of ccl4 are needed to have 5.000 mol?
769.0 grams
