Titration calculations - working out cation
You must have forgotten a factor $2$ somewhere.
You are right is thinking that $24.10$ mL is needed for the final titration. But now let's redo the calculations.
$24.10$ mL $\ce{NaOH} ~0.25$ M contains $24.10·0.25 = 6.025$ millimole $\ce{NaOH}$. Obviously this $\ce{NaOH}$ has neutralized the same amount $6.025$ millimole $\ce{HCl}$. As a consequence, there was $10$ times more $\ce{HCl}$ in the $250$ mL flask. This is $0.06025$ mol $\ce{HCl}$, and was the residual amount of $\ce{HCl}$ remaining after reaction of HCl plus the unknown carbonate. It also means that after adding $0.100$ mole $\ce{HCl}$ in the initial flask, the difference $0.100 - 0.06025 = 0.03975$ mol $\ce{HCl}$ had been used to destroy the carbonate.
But whatever the formula of the initial carbonate, $2~\ce{H+}$ are needed to react with  $1~ \ce{CO3^{2-}}$. So the amount of unknown $\ce{XCO3}$ or $\ce{Y2CO3}$ carbonate in the original powder is half the amount of $\ce{HCl}$ and it is : $0.03975/2$ mol = $0.019875$ mol $\ce{XCO3}$.
As the mass of this carbonate is $2.50$ g, its molar mass is: M = m/n = $2.50/0.019875 = 125.8$ g/mol. As the molar mass of $\ce{CO3}$ is $60$ g/mol, it means that the molar mass of $\ce{X}$ is $125.8 - 60 = 65.8$ g/mol.
This final value may be twice the atomic mass of a monovalent cation, or the atomic mass of a bivalent cation. As there are no monovalent cation with an atomic mass of about $32.9$, the unknown element must be a bivalent cation. And the only element with such an atomic mass is Zinc : A(Zn) = $65.39$
So the unknown carbonate is $\ce{ZnCO3}$