Question

Suppose the fluid in the stomach of a man suffering from indigestion can be considered to be 200.ml of a 0.033 m hcl solution. what mass of nahco3 would he need to ingest to neutralize this much hcl

Answer 1

HCl + NaHCO3 = NaCl + H2O + CO2

We have 0.033*0.2 = 0.0066 mol of Hcl

According to reaction we need the same amount of NaHCO3
M(NaHCO3) = 23+1+12+48=84g/mold
m = 0.0066mol * 84g/mol = 0.5544g