What volume of .100 m hcl is required to titrate .100 g of sodium bicarbonate to its equivalence point?

The reaction is:  HCl + Na HCO3 → NaCl + H2CO3, which is already balanced.

Then the molar ratios are 1 mol HCL : 1 mol NaHCO3

You know the mass of NaHCO3 is 0.100 g, then you can find the number of moles using the molar mass of NaHCO3

molar mass of NaHCO3 is 23 g/mol + 1 g/mol + 12g/mol + 3*16g/mol = 84 g/mol

And the number of moles is: 0,100 g / 84 g/mol  = 0,0012 mol of NaHCO3.

Then, from the ratio 1:1, you know that, at the titration point, the number of moles of HCl is the same: 0,0012 mol of HCl.

Now from the fomula of molarity you have: M = #of moles / V in liters

=> V in liters = # of moles / M = 0.0012 mol / 0.100 M = 0.012 liters

0.012 liters of HCL