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The amount of substance that will dissolve depends on the temperature and the solvent used. Thus, there is no way to tell how much will dissolve unless such a test is actually carried out (since that concentration of acid is not a standard solvent used in such experiments).
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What is purpose of adding HCl to the CaCO3 -H2O mixture prior to dilute?
It is the only way of dissolving CaCO3. HCl + CaCO3 --> Ca2+ + H2O + CO2 In neutral water CaCO3 is insoluble.
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 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.
What products do you get when CaCl2 plus NaHCO3?
Nacl+Hcl+CaCO3
What is the equation for calcium carbonate with hydrochloric acid?
You can find a answer fromCalcium carbonate - Wikipedia
What is purpose of adding HCl to the CaCO3 -H2O mixture prior to dilute?
It is the only way of dissolving CaCO3. HCl + CaCO3 --> Ca2+ + H2O + CO2 In neutral water CaCO3 is insoluble.
If you have the chemical equation caco3 and hcl equal cacl3 etc. are caco3 and hcl particles molecules or ions?
They are molecules.
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
What is the Balanced equation for calcium carbonate and hydrochloric acid reacting?
CaCO3 + HCl --> CaHCO3 or with excess of HCl CaCO3 + 2HCl --> CaCl2 +CO2 + H2O
How many moles of HCl can 0.055 g of calcium carbonate neutralize?
First write the balanced chemical equation for this reaction CaCO3+2HCl-->CaCl2+H2O+CO2 It is clear from this equation that one mol of HCl can react with half a mol of CaCO3. Thus, one half of .025 mols CaCO3 will react with .025 mols HCl. That makes .0125 mols CaCO3.
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.
What products do you get when CaCl2 plus NaHCO3?
Nacl+Hcl+CaCO3
What is the equation for calcium carbonate with hydrochloric acid?
You can find a answer fromCalcium carbonate - Wikipedia
How many grams of CaCl2 can be obtained if 14.6 g HCl is allowed to react with excess CaCO3?
22.2g CaCl2
What is the word formula for calcium carbonate and hydrochloric acid?
CaCO3 AND HCl
How many moles are dissolved in 200 ml of 0.100 M HCl?
20
What is the balancing equation of hcl plus caco3--cacl2 plus h20 plus CO2?
CaCO3+2HCl----->CaCl2+H2O+CO2
