How to find the concentration of a titrant

"for full credit use a systematic approach to show all calculations required to get from the experiemental data to the concentration of the HCl soution"

initial burette reading: 50.00mL
final reading 31.25mL
volume delivered: 18.75mL
mass of sodium carbonate added to 100.00mL volumetric flask: 0.040g
volume of sodium carbonate solution titrated: 25.00mL

find the concentration of a titrant?
"for full credit use a systematic approach to show all calculations required to get from the experiemental data to the concentration of the HCl soution"

initial burette reading: 50.00mL
final reading 31.25mL
volume delivered: 18.75mL
mass of sodium carbonate added to 100.00mL volumetric flask: 0.040g
volume of sodium carbonate solution titrated: 25.00mL

Titration Equation:
Liters of acid * Normality of acid = Liters of base * Normality of base

Concentration of Na2CO3 = moles ÷ liters
Sodium carbonate = Na2CO3
Mass of 1 mole = (2 * 23) + 12 + (3 *16) = 106 grams
Moles of Na2CO3 in 100 ml = 0.04 ÷ 106
Molarity of Na2CO3 = (0.04 ÷ 106) ÷ 0.100
Normality of Na2CO3 = [(0.04 ÷ 106) ÷ 0.100] * 2
0.01875 * Normality of acid = 0.025 * [(0.04 ÷ 106) ÷ 0.100] * 2

Normality of acid = 0.010062893

OR

Balanced reaction equation:
Na2CO3 + 2 HCl → 2 NaCl + H2CO3 (H2O + CO2)
According to the balanced equation above, 1 mole of Na2CO3 reacts with 2 moles of HCl to produce 2 moles of NaCl and 1 mole of H2CO3.

Sodium carbonate = Na2CO3
Mass of 1 mole = (2 * 23) + 12 + (3 *16) = 106 grams
Moles of Na2CO3 in 100 ml = 0.04 ÷ 106
Moles of Na2CO3 in 25 ml = ¼ * 0.04 ÷ 106 = 0.01 ÷ 106

1 mole of Na2CO3 reacts with 2 moles of HCl, so (0.01 ÷ 106) moles of Na2CO3 reacts with 2 * (0.01 ÷ 106) moles of HCl

Volume of HCl = 18.75mL = 0.01875 liters
Concentration of HCl = (0.02 ÷ 106) ÷ 0.01875 = 0.010062893
The HCl solution is approximately 0.01 M