What volume of 0.200 m hcl is required for the complete neutralization of 2.00 g of nahco3 (sodium bicarbonate)?

The reaction between HCl and NaHCO3 will be;
HCl + NaHCO3 = NaCl + CO2 + H2O
The relative formula mass of NaHCO3 is 84 g/mol
Moles of NaHCO3 in 2 g will be; 2/84 =0.0238 moles
The mole ratio of HCl and NaHCO3 is 1;1
Thus, the number of moles of HCl is 0.0238 moles
 The volume of HCl will be;
  = 0.0238 moles/0.2 
  = 0.119 liters or 119 cm³

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