If 2.34 g of nacl was formed how many moles of nahco3 must have been used in the reaction? (report only the numerical portion of your answer [i.
e. leave off the units] to 3 significant digits)

M(NaCl)=2.34 g
n(NaHCO3)=?

NaHCO3 + HCl = NaCl + H2O + CO2

According to the reaction equation, it can be seen that NaHCO3 and NaCl have following stoichiometric ratio: 

n(NaHCO3) : n(NaCl) = 1 : 1

n(NaHCO3) = n(NaCl) = m/M = 2.34/58=0.04

So, when moles are known, we can calculate mas of NaHCO3:

m(NaHCO3) = nxM = 0.04 x 84 = 3.36