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21 May, 22:13

How many grams of MgCO3 are required to neutralize 200. mL of stomach acid HCl, which is equivalent to 0.0465 MHCl?

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  1. 21 May, 22:19
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    0.392 g

    Explanation:

    We are given the following;

    Volume of HCl is 200 mL

    Molarity of HCl is 0.0465M

    We are required to calculate the mass of MgCO₃ required

    Step 1: Write the balanced equation for the reaction The balanced equation for the reaction is;

    MgCO₃ (s) + 2HCl (aq) → MgCl₂ (aq) + CO₂ (g) + H₂O (l)

    Step 2: Calculate the moles of HCl

    When given the molarity of a compound and the volume, the number of moles can be calculated by;

    Number of moles = Molarity * Volume

    Therefore;

    Volume of HCl = 0.0465 M * 0.2 L

    = 0.0093 moles

    Step 3: Calculating the number of moles of MgCO₃

    From the equation, one mole of MgCO₃ reacts with two moles of HCl

    Therefore, the mole ratio of MgCO₃ : HCl is 1 : 2

    Hence, moles of MgCO₃ = Moles of HCl : 2

    = 0.0093 moles : 2

    = 0.00465 moles

    Step 4: Mass of MgCO₃

    To calculate the mass of a compound we need to multiply the molar mass of a compound with the number of moles.

    Molar mass of MgCO₃ is 84.314 g/mol

    Thus, Mass of MgCO₃ = 0.00465 moles * 84.314 g/mol

    = 0.392 g

    Therefore, 0.392 g of MgCO₃ are required to neutralize the acid.
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