HCl = 0.1 M.
find grams of Acid to neutralize the tablet.
n = 0.00739 mol of CO3-2.
D = 1.02 g/ml.
Find the moles of CO3-2.
CO3-2 + 2HCl <---> H2CO3 + 2Cl-.
therefore,
we need 2 mol of CO3-2 to neutralize 1 mol of HCl.
since we have 0.00739 mol of CO3-2, that will neutralize 2X mol of HCl.
2*0.00739 = 0.01478 mol of HCl will be neutralized.
Now, find the mass of Acid :
M = mol/V,
V = mol/M.
Substitute data of acid :
V = 0.01478 mol / 0.1 M = 0.1478 Liters of stomach acid
or 147.8 ml of stomach acid
now find mass:
D = mass/V.
mass = D*V = 1.02*147.8 = 150.76 grams of Stomach Acid will be neutralized
