Question

How many milliliters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3 (= 84.02 g/mol)?

HCl(aq) + NaHCO3(s) ? NaCl(s) + H2O(l) + CO2(g)
a. 175 mL
b. 536 mL
c. 276 mL
d. 572 mL
e. 638 mL

Answer 1

Answer: a (175 mL)
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