Help with Neutralization Power of an Antacid question....?

Question

Assuming that stomach acid is 0.150 M HCl (p = 1.00 g/ml) how much mass of acid would a 500 mg tablet composed of Na2CO3 neutralize?

How would you begin to figure this out......what equation would you use? I feel stupid, but I just can't figure out how to start with solving this? Any advice would be appreciated! Thank you!

Answer 1

2HCl + Na2CO3 = 2NaCl + CO2 + H2O

Assuming that the tablet contains only Na2CO3 , then

mol Na2CO3 = 0.5 g/ (2(23) + 12+ 3(16)) = 0.00472 mol

According to the stoichiometric eqn. above 2 mol HCl scompletely reacts with 1 mol Na2CO3

mol HCl reacted = 2*0.00472 = 0.00944 mol

volume of HCl reacted = 0.00944 /  0.150 = 0.0629 L =62.9 mL

But you were given the density of HCl to be 1 g/ mL

therefore mass of HCl = 62.9 g