Final answer:
To neutralize 14.2 mL of 0.15 M HCl, 0.00213 moles of KOH are required. Since KOH has a molar mass of 56.11 g/mol, this corresponds to 0.11945 grams. The closest answer is 0.11 grams, option b.
Step-by-step explanation:
To determine how many grams of KOH are needed to neutralize 14.2 mL of 0.15 M HCl in stomach acid, we first need to calculate the number of moles of HCl. This can be done using the formula: moles of HCl = volume in liters × molarity (M). Given that 14.2 mL is equivalent to 0.0142 liters, we can determine moles of HCl: 0.0142 L × 0.15 M = 0.00213 moles of HCl.
Since KOH and HCl neutralize each other in a 1:1 molar ratio, the same number of moles of KOH are needed to neutralize the HCl. The molar mass of KOH is approximately 56.11 g/mol, so the mass of KOH needed is 0.00213 moles × 56.11 g/mol = 0.11945 grams.
Therefore, the closest answer to the amount of KOH needed to neutralize the HCl is option b. 0.11 grams.