Question

How many mL of stomach acid (assume it is .035 M HCI) can be neutralized by an antacid tablet containing .231g

Mg(OH)2?

Answer 1

Mg(OH)2(aq.) + 2HCl(aq.) --> MgCl2(aq.) + 2H2O(l)

n(HCl)/n(Mg(OH)2)= 2/1

n(HCl)= 2*n(Mg(OH)2)

c= n/V; n= c*V

n= m/M

c(HCl)*V(HCl)= 2*m(Mg(OH)2)/M(Mg(OH)2)

V(HCl)= 2*m(Mg(OH)2)/(M(Mg(OH)2)*c(HCl))= 2 * 0,231 g / (58,326 g mole^-1 * 0,035 mole dm^-3)= 0,226 dm^3= 226 cm^3 (cm^3= mL)