Which out of dil. `H_2SO_4`, HCl and `HNO_3` can be used in the preparation of carbon dioxide from `PbCO_3` ?

Explanation

`PbCO_3` reacts with dilute `H_2SO_4` and dilute HCl to form insoluble `PbSO_4 and PbCl_2` respectively. An insoluble protective coating of the salt is formed over the surface of the marble pieces and this cuts off the contact between the acid and marble. So the reaction stops and no more `CO_2` is produced. It is for this reason, dilute`H_2SO_4` and HCl cannot be used in the preparation of `CO_2` from `PbCO_3`. When dilute `HNO_3` is used, highly soluble `Pb(NO_3)_2` s formed and so, the evolution of `CO_2` takes place smoothly . Therefore, dilute `HNO_3` can be used in the preparation of `CO_2` from `PbCO_3`.
`PbCO_3 + 2HCl to underset("Insoluble coating")(2PbCl_2darr) + CO_2uarr + H_2O`
`PbCO_3 + H_2SO_4 to underset("insoluble coating")(PbSO_4darr) + CO_2uarr + H_2O`
`PbCO_3 + 2HNO_3 to underset("soluble")(Pb(NO_3)_2) + CO_2uarr + H_2O`