Question

A tablet contains the same number of moles of CaCO3 and MgCO3. One tablet reacted with excess hydrochloric acid to produce 0.44g of Carbon Dioxide. CaCO3 + 2HCl = CaCl2 + CO2 + H2O MgCO3 + 2HCl = MgCl2 + CO2 + H2O How many moles of CaCO3 are in a tablet?

I know that in CO2 there are 0.01 moles. I don't understand how to get to CaCO3. Answers would be appreciated.

Answer 1

We interrogate the reaction.....

`MCO_3(s) + 2HCl(aq) rarr MCl_2(aq) + CO_2(g)uarr + H_2O(l)`

Explanation:

And thus since we know the molar quantity of carbon evolved, we know that this molar quantity represents the moles of each starting metal carbonate given the boundary conditions of the problem. And the moles of carbon dioxide is HALF a molar equivalent of the hydrogen chloride added....

`"Moles of carbon dioxide"-=(0.44*g)/(44.01*g*mol^-1)=1.0xx10^-2*mol`

And thus there were `5.0xx10^-3*mol` with respect to `MgCO_3`, and `5.0xx10^-3*mol` with respect to `CaCO_3`.

And there were a mass of `5.0xx10^-3*molxx100.09*g*mol^-1=0.50*g` with respect to calcium carbonate.... Happy?