Apr 2, 2018 · The question is: How many g of 30% HCl-acid is needed in the beginning. $$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$ So for each dissolved molecule of $\ce{CaCO3}$: 2 molecules of $\ce{HCl}$ are used; 1 molecule of $\ce{H2O}$ is produced; 1 molecule of $\ce{CaCl2}$ is produced; 1 molecule of $\ce{CO2}$ is …

Feb 28, 2016 · I have a wonderful reaction of marble chips, $\ce{CaCO3}$, with hydrochloric acid, $\ce{HCl}$, and carbon dioxide was released beautifully (fast, large volume, easy to measure and makes good visual effect too). But there is no reaction between $\ce{CaCO3}$ and $\ce{H2SO4}$. Why not?

Nov 10, 2013 · The reaction between CaCO3 and HCl is as follows:CaCO3 + 2HCl --> CaCl2 + CO2 +H2O Therefore, molar ratio between CaCO3 and HCl is 1:2. 0.042 moles HCl reacted, therefore 0.021 mol CaCO3 were present in the sample. Multiplying this by the molar mass yields 0.021mol * (40+12+48)g/mol = 2.1 grams CaCO3.

Aug 14, 2023 · We add 0.2 grams of CaCO3 to 100ml of 0.1M HCl. The 0.1M HCl is dissolved in pure water. This should generated 0.002M of CO2, and if I did my calculations right that should be about 51.02ml of gas. But everytime we measure we get around 30% less.

Aug 26, 2019 · I conducted an experiment in which $\\pu{10mL}$ of dilute hydrochloric acid was added to a beaker, and this acid was heated up to room temperature ($\\pu{22 ^\\circ C}$) using a Bunsen burner. Then, $...

Mar 14, 2017 · You could add hydrochloric acid again to convert it into soluble Calcium chloride, according to equilibrium: $$\ce{CaCO3 (s) + 2 HCl (aq) <=> CaCl2 (aq) + H2CO3(aq)}$$ You could use carbonic acid instead $$\ce{CaCO3 (s) + H2CO3 <=> Ca(HCO3)2}$$ this is the reason shells are getting dissolved by an increase of the amount of carbon dioxyde in ...

Oct 5, 2014 · The lattice energy (LE), the energy of formation of the crystal between infinitely separated ions. As LE is proportional to the charges of its ions, for calcium carbonate, its LE would be roughly 4× the LE of sodium chloride. $$\ce{CaCO3 (s) -> Ca^2+ (g) + CO3^2- (g)}$$ This value is always positive, as energy is required to separate the ions.

Jun 16, 2016 · $15.0~\mathrm{mL}$ of $1.4~\mathrm{M}\ \ce{HCl}$ was mixed with $1.00~\mathrm{g}$ of limestone (impure $\ce{CaCO3}$) until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $200~\mathrm{mL}$ with water.

Jan 10, 2016 · The question is a bit poorly written. But it also gives a sense to the procedure of dissolution of chalk upon treatment with $\ce{HCl}$ and dilution of the solution. The reaction that takes place after the treatment with $\ce{HCl}$ is; $\ce{CaCO3}$+ 2$\ce{HCl}$ → …

Jan 4, 2016 · We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles.