How many milliliters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3 according to the following reaction? HCl (aq) + NaHCO3 (s) → NaCl (s) + H2O (l) + CO2 (g)

Answer 1

HCl (aq) + NaHCO3 (s) → NaCl (s) + H2O (l) + CO2 (g)

23.2 g NaHCO3 x 1 mol NaHCO3/84.007 g = 0.276 moles

moles HCl needed = 0.276 mol NaHCO3 x 1 mol HCl/mol NaHCO3 = 0.276 moles HCl needed

Volume HCl: (x L)(1.58 mol/L) = 0.276 moles

x = 0.175 L = 175 mls