<h3>
Answer:</h3>
0.392 g
<h3>
Explanation:</h3>
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
<h3>Step 1: Write the balanced equation for the reaction </h3>
- The balanced equation for the reaction is;
MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + CO₂(g) + H₂O(l)
<h3>Step 2: Calculate the moles of HCl</h3>
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
<h3>Step 3: Calculating the number of moles of MgCO₃</h3>
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
<h3>Step 4: Mass of MgCO₃</h3>
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.
