You pipetted 15.0mL of Na2CO3(aq). You then added 20.0mL of Sr(NO3)2(aq). Your dry filter paper was 1.69g. You filter paper and precipitate is 3.76g. What is the concentration of the Na2CO3(aq)?
Very little information. Do you want the concentration of Na2CO3 before or after the reaction. What is the concn of Na2CO3 initially? What is the concn of Sr(NO3)2 initially? What units of concn do you want? I will assume g/L for your answer and I will assume you had equal mols Na2CO3 and Sr(NO3)2 so the reaction is Na2CO3(aq) + Sr(NO3)2(aq) ==> SrCO3(s) + 2NaNO3(aq)
IF mols Na2CO3 and mols Sr(NO3)2 at the beginning were equal, then the ppt of SrCO3 is 3.76-1.69 = 2.07 g and you had neither of the reactants left. You must have started with
2.07 g SrCO3 x (molar mass Na2CO3/molar mass Na2CO3) = ? grams Na2CO3 and that dissolved in 35 mL gives you the concentration of Na2CO3 in g/mL.