Assuming stomach acid is 0.1 m hcl, how many grams of stomach acid would have been neutralized by your tablet (use the average value)?

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