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