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3 March, 11:18

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) ?

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  1. 3 March, 14:59
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    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
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