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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 equation for calcium carbonate with hydrochloric acid?
You can find a answer fromCalcium carbonate - Wikipedia
If a few drops of HCl spilled over into the CaCO3 sample error in the reported percent CaCO3 will result from this poor laboratory technique?
Yes, if HCl (hydrochloric acid) spilled into the CaCO3 sample, a chemical reaction will occur. This will alter the mass of CaCO3 present in the sample, leading to an error in the reported percent CaCO3 due to the loss of some CaCO3 in the reaction with HCl.
If 1.25 grams of pure CaCO3 required 25.5 mL of a HCl solution for complete reation what is the molarity for the HCl solution?
When calcium carbonate (CaCO3) reacts with hydrochloric acid (HCl) the reaction is:- CaCO3 + 2 HCl -> CaCl2 + H2O + CO2 The weight for 1 mole of calcium carbonate is the atomic weight in grams or 40.8 + 12.011 + 3 x 15.9994 = 100.8092 grams 1.25 grams of CaCO3 is therefore 1.25/100.8092 or .0124 mole There must have been twice this many moles of HCl in the solution or 0.0248 mole. The molarity of the solution is the number of moles per litre so with 0.0248 in 25.5 ml there must be .97 mole in 1 litre. The HCl solution must therefore be .97 molar
How do you complete molecular equation for hcl and caco3?
The molecular equation for the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3) is: HCl + CaCO3 → CaCl2 + CO2 + H2O. This reaction produces calcium chloride (CaCl2), carbon dioxide (CO2), and water (H2O) as products.
If you have the chemical equation caco3 and hcl equal cacl3 etc. are caco3 and hcl particles molecules or ions?
They are molecules.
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
What is the word formula for calcium carbonate and hydrochloric acid?
Calcium carbonate and Hydrochloric acid = Calcium chloride, water and carbon dioxide. Here is the BALANCED reaction equation. CaCO3(s) + 2HCl(aq) = CaCl2(aq) + H2O(l) + CO2(g)
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 equation for calcium carbonate with hydrochloric acid?
You can find a answer fromCalcium carbonate - Wikipedia
If a few drops of HCl spilled over into the CaCO3 sample error in the reported percent CaCO3 will result from this poor laboratory technique?
Yes, if HCl (hydrochloric acid) spilled into the CaCO3 sample, a chemical reaction will occur. This will alter the mass of CaCO3 present in the sample, leading to an error in the reported percent CaCO3 due to the loss of some CaCO3 in the reaction with HCl.
If 1.25 grams of pure CaCO3 required 25.5 mL of a HCl solution for complete reation what is the molarity for the HCl solution?
When calcium carbonate (CaCO3) reacts with hydrochloric acid (HCl) the reaction is:- CaCO3 + 2 HCl -> CaCl2 + H2O + CO2 The weight for 1 mole of calcium carbonate is the atomic weight in grams or 40.8 + 12.011 + 3 x 15.9994 = 100.8092 grams 1.25 grams of CaCO3 is therefore 1.25/100.8092 or .0124 mole There must have been twice this many moles of HCl in the solution or 0.0248 mole. The molarity of the solution is the number of moles per litre so with 0.0248 in 25.5 ml there must be .97 mole in 1 litre. The HCl solution must therefore be .97 molar
How do you complete molecular equation for hcl and caco3?
The molecular equation for the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3) is: HCl + CaCO3 → CaCl2 + CO2 + H2O. This reaction produces calcium chloride (CaCl2), carbon dioxide (CO2), and water (H2O) as products.
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.
What mass of calcium carbonate in grams can be dissolved by 3.9g of HCl?
To solve this, you have to be aware that this is a acid-base reaction, and that HCl is a gas, that usually is not applied in this form.However: Hydrochloric acid reacts with the calcium salt of carbonic acid, to form calcium chloride, water and (volatile) carbon dioxide.Thus, you must first calculate the moles (n) of H+ contained in 3.9g of HCl. 1 mole of HCl contains 1 mole of H+. So you can calculate:M(HCl) = M(H) + M(Cl) = 36.45g/molm(HCl) = 3.90gn(HCl) = m(HCl) / M(HCl) = 0.11molNext, you must calculate the moles of carbonate that can be dissolved.Using the following formula:CO32- + 2 H+ ↔ H2O + CO2↑you can see, that you need 2 moles of H+ for 1 mole of CO32-.Subsequently, you have to calculate the molar mass of calcium carbonate:M(CaCO3) = 40.08g/mol + 12.01g/mol + 3*16.00g/mol = 100.09g/molAnd finally, you can calculate the mass of calcium carbonate you can dissolve using 0.11mol HCL:m(CaCO3)= M(CaCO3) * [½ * n(HCl)] = 5.35gFrom the equations above, considering the molarities, we can draw a more dense formula that allows us to neglect the absolute molarities, so we only have to use the relative molarities. The equation can also be used to check if we calculated correctly)m(A)/(2*M(A)) = m(B)/M(B)We transpose to calculate m(B):m(B)= (½*m(A)/M(A))*M(B)and when we insert the values:m(CaCO3) = (0.50*(3.90g/36.45g/mol))*100.09g/mol = 5.35gAnd next time, you'll be able to do this by yourself ;)
What is the balancing equation of hcl plus caco3--cacl2 plus h20 plus CO2?
The balanced equation for the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3) is: 2HCl + CaCO3 -> CaCl2 + H2O + CO2.
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.
