Calculate the weight percent of limestone in a rock sample
The question is a bit misleading ;)

Your calculation of the moles of $\ce{CaCO3}$ would be correct if the rock sample would be pure 100% calcium carbonate. But then, all the titration would be a waste of time: There's 100 wt-% $\ce{CaCO3}$ in pure $\ce{CaCO3}$.

So, there's apparently something else in the rock sample (that doesn't disturb the titration).

However, what you know for sure is: 


the stoichiometry of the reaction between $\ce{CaCO3}$ and $\ce{HCl}$. With other words: How many moles of $\ce{HCl}$ are needed to react one mole of $\ce{CaCO3}$
concentration and volume of the $\ce{HCl}$ used in excess to neutralize all the $\ce{CaCO3}$
concentration and volume of the $\ce{NaOH}$ to neutralize all the $\ce{HCl}$ that is left after reacting all $\ce{CaCO3}$




HINT: 

Did you consider to solve the back titration experiment by starting with the second neutralisation step? 


   If 11.56mL of a 1.010 M $\ce{NaOH}$ was used to neutralize the remaining $\ce{HCl}$, how much was left of it? Compare that with the initial amount...