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: 


$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$


Then:


$$\ce{HCl + NaOH -> H2O + NaCl}$$


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 $\ce{CaCO3}$




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**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...
> 
> 
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 2

1. Calculate the amount of excess acid:


$$\mathrm{x mol HCl = 11.56mL NaOH \times \frac{1.010 mol NaOH}{1000mL}\times \frac{1 mol HCl}{1mol NaOH} = 0.01167 mol HCl }$$


2. Determine the amount of acid originally added:


$$\mathrm{x mol HCl = 30 mL HCl \times \frac{0.035 mol NaOH}{1000mL}= 0.0011 mol HCl }$$


3. Realize more acid was neutralized than was originally added. There must be an error in the question some where.


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