I titrated 0.25 grams of soda ash with 43.7 mL of HCl (0.115M) and 25 mL of H2O. I am getting over 100% sodium carbonate in my answer here are my steps that I took to get the answer can somebody help me see what I did wrong?

mols HCl=(0.115M)x (0.0437L)=0.005 mol HCl

1/2 x (0.005)= 0.0025mol Na2CO3

0.0025mol x 105.99g= 0.42g Na2CO3

0.42g/0.25g x 100%= 168%

To analyze the error in your calculation, let's retrace the steps and identify the mistakes:

1. Calculation of moles of HCl: You correctly calculated the moles of HCl using the concentration and volume. Therefore, this step is correct.

2. Calculation of moles of sodium carbonate (Na2CO3): The balanced chemical equation for the reaction between HCl and Na2CO3 is:

2HCl + Na2CO3 -> 2NaCl + CO2 + H2O

Since the stoichiometry of the reaction is 2:1 (2 moles of HCl react with 1 mole of Na2CO3), you need to divide the moles of HCl by 2 (not multiply by 1/2) to find the moles of Na2CO3.

Correct calculation:
0.005 mol HCl / 2 = 0.0025 mol Na2CO3

3. Calculation of mass of Na2CO3: You correctly multiplied the moles of Na2CO3 by the molar mass of Na2CO3 (105.99 g/mol) to find the mass.

Correct calculation:
0.0025 mol Na2CO3 x 105.99 g/mol = 0.265 g Na2CO3

4. Calculation of percentage purity: The percentage purity is calculated by dividing the actual mass of the compound (0.265 g) by the expected mass (0.25 g) and multiplying by 100%.

Correct calculation:
(0.265 g / 0.25 g) x 100% = 106%

Therefore, the correct percentage purity of your soda ash sample is 106%. The result indicates that your soda ash sample is slightly impure, not over 100% pure as you initially thought.