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2 HCl + Na₂CO₃ → 2 NaCl + H₂O + CO₂
Mole ratio HCl : Na₂CO₃ = 2 : 1

No. of moles of Na₂CO₃ = (4.3059 g) / (106 g/mol) = 0.04062 mol
No. of moles of HCl = (0.04062 mol) × 2 = 0.08124 mol
Molarity of HCl = (0.08124 mol) / (46.79/1000 L) = 1.736 mol/L

2 HCl + Na2CO3 → 2 NaCl + H2O + CO2

According to the balanced equation above, 2 moles of hydrochloric acid reacts with one mole of sodium carbonate. The first step is to determine the number of moles of sodium carbonate

n = 4.3059 ÷ 106

This is approximately 0.00406 mole. The number of moles of hydrochloric acid is twice this amount.

n = 8.6118 ÷ 106

Number moles = Volume in liters * molarity
8.6118 ÷ 106 = 0.04679* molarity
M = 8.6118 ÷ 4.95974

This is approximately 1.736 moles per liter.