<h3>
Answer:</h3>
0.1252 g of HCl
<h3>
Explanation:</h3>
Antacid acts to neutralize excessive acid in the stomach.
The stomach contains hydrochloric acid, HCl.
Therefore; a reaction of antacid (magnesium hydroxide) and HCl will be;
Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) +2 H₂O(l)
We are given;
Mass of magnesium hydroxide as 0.10 g
We can calculate the mass of HCl neutralized by the acid using the following simple steps;
Step 1 : Number of moles of Mg(OH)₂
Number of moles is calculated by dividing the mass of a compound with the relative formula mass of the compound.
Relative formula mass of the compound = 58.32 g/mol
Therefore;
Number of moles of Mg(OH)₂ = 0.10 g ÷ 58.32 g/mol
= 0.001715 mol
Step 2: Number of moles of HCl neutralized by 0.10 g of Mg(OH)₂
Moles of Mg(OH)₂ = 0.001715 mol
But, from the equation 1 mole of Mg(OH)₂ reacts with 2 moles of HCl
Thus;
0.001715 moles Mg(OH)₂ will neutralize;
= 0.001715 × 2
= 0.003430 moles HCl
Step 3: Mass of HCl neutralized
Mass is calculated by multiplying the number of moles by the relative formula mass of a compound.
Molar mass of HCl = 36.5 g/mol
Therefore;
Mass = 0.003430 mol × 36.5 g/mol
= 0.1252 g HCl
Therefore; 0.10 g of Mg(OH)₂ neutralizes 0.1252 g of HCl
