Calculate the weight percent of limestone in a rock sample A 1.245 g sample of limestone ($\ce{CaCO3}$) is pulverized and then treated with 30 mL of 0.035 M $\ce{HCl}$. The excess acid requires 11.56 mL of 1.010 M $\ce{NaOH}$ to be neutralized. Calculate the %wt of limestone in the rock.

The first balanced equation would be: 

$$\ce{CaCO3 + 2HCl -&gt; CaCl2 + H2O + CO2}$$

Then:

$$\ce{HCl + NaOH -&gt; H2O + NaCl}$$

I figured out the moles of $\ce{CaCO3}$ if that's useful for this problem 1.245 g of $\ce{CaCO3}$ / 100.089 g (molar mass)= 0.0124 M

I have this but I don't know the next steps on how to solve the problem.
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...