Question

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

Answer 1

Answer:

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.