How many grams or moles of pure HCl are needed to completely react with 0.581 grams of Na2CO3?

Molar mass of Na₂CO₃ = (23.0×2 + 12.0 + 16.0×3) g/mol = 106.0 g/mol
No. of moles of Na₂CO₃ reacted = (0.581 g) / (106.0 g/mol) = 0.00548 mol

2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂
Mole ratio HCl : Na₂CO₃ = 2 : 1
Hence, no. of moles of HCl needed = (0.00548 mol) × 2 = 0.01096 mol ≈ 0.0110 mol

Molar mass of HCl = (1.0 + 35.5) g/mol = 36.5 g/mol
Mass of HCl needed = (0.01096 mol) × (36.5 g/mol) = 0.400 g

If the expected products include NaCl as the only repository of sodium and chlorine, you need twice as many moles of HCl as you've got of Na2CO3. The molar mass of Na2CO3 would be 46 + 12 + 48 = 106, so the 0.581 grams is 5.48 millimoles. Therefore, you need 11.0 millimoles of HCl (which is 0.400 grams).

Na2O3+2HCL=2NaCl+H2O+O2
molar mass of Na2Co3 is 105.8
molar mass of HCl is 36.45
from gram of Na2Co3 to moles of NA2CO3
0.581g Na2CO3 *1mole Na2CO3/105.8g Na2CO3*2moleHCl/1 mole Na2CO3=0.01089, to correct sig fig= 0.0110
now from mole of HCL to Gram of HCl= (but do not use the rounded number)
0.01098 moles of HCl*36.45g of HCl/1 mole of HCl= 0.400 g of HCL:)))