What mass of calcium carbonate in grams can be dissolved by 3.9g of HCl?

Answer 1

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/mol

m(HCl) = 3.90g

n(HCl) = m(HCl) / M(HCl) = 0.11mol

Next, 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/mol

And finally, you can calculate the mass of calcium carbonate you can dissolve using 0.11mol HCL:

m(CaCO3)= M(CaCO3) * [½ * n(HCl)] = 5.35g

From 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 = 5.35g

And next time, you'll be able to do this by yourself ;)

Answer 2

CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

6.00 grams HCl (1 mole HCl/36.458 grams)(1 mole CaCO3/2 mole HCl)(100.09 grams/1 mole CaCO3)

= 8.24 grams CaCO3

seems unreasonable, check answer if you can