Final answer:
To neutralize 25 mL of 0.10 M HCl, 0.0729 g of Mg(OH)2 is required. This calculation is based on stoichiometry and the molar masses involved in the neutralization reaction.
Step-by-step explanation:
To determine the mass of Mg(OH)2 required to neutralize 25 mL of stomach acid, which is 0.10 M HCl, we need to use the concept of stoichiometry from the balanced chemical equation of the neutralization reaction:
Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O
First, we calculate the moles of HCl in 25 mL of the given solution:
0.025 L * 0.10 M = 0.0025 mol of HCl
Since it takes 1 mole of Mg(OH)2 to neutralize 2 moles of HCl, the amount of Mg(OH)2 required is half the amount of HCl:
0.0025 mol HCl * (1 mol Mg(OH)2 / 2 mol HCl) = 0.00125 mol of Mg(OH)2
The molar mass of Mg(OH)2 is approximately 58.32 g/mol, so:
0.00125 mol * 58.32 g/mol = 0.0729 g of Mg(OH)2