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