Respuesta :
0.06642 grams of KOH are needed to neutralize 13.4 mL of 0.17 M HCl in stomach acid.
Define the molarity of a solution.
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
Step 1: Data given
Volume of stomach acid (HCl) = 13.4 mL =0.0134 L
Molarity HCl = 0.17 M
Step 2: The balanced equation
2KOH + 2 HCl → 2KCl + 2H2O
For 2 mol of KOH, we need 2 moles of HCl to produce 2 mol of KCl and 2 moles of H2O
Step 3: Calculate moles of HCl
Moles HCl = molarity x volume
Moles HCl = 0.17 M x 0.0134 L
Moles HCl = 0.002278 moles
Step 4: Calculate moles of KOH
For2 mol of KOH we need 2 moles of HCl to produce 2 mol of KCl and 2 moles of H2O
For 0.00025 moles of HCl we need 0.002278/2 = 0.001139 moles of KOH
Step 5: Calculate mass of KOH
Mass of KOH = moles x molar mass
Mass of KOH = 0.001139 moles x 58.32 g/mol
Mass of KOH = 0.06642 grams
Hence, 0.06642 grams of KOH are needed to neutralize 13.4 mL of 0.17 M HCl in stomach acid.
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