How many grams of mg oh 2 will be needed to neutralize 25 ml of stomach acid of stomach acid is 0.10 m Hcl?
To neutralize the 0.10 M HCl in 25 mL, we first calculate the moles of HCl present: 0.10 mol/L * 0.025 L = 0.0025 mol HCl. Since 1 mole of HCl reacts with 2 moles of Mg(OH)2 (according to the balanced chemical equation), we need 0.0050 moles of Mg(OH)2. The molar mass of Mg(OH)2 is 58.3 g/mol, so we need 0.0050 mol * 58.3 g/mol ≈ 0.29 grams of Mg(OH)2.