Calculate the normality of a potassium hydroxide solution if 27.7 mL of the solution is needed to neutralize 0.477 g of sulfamic acid in the reaction NH₂SO₃H + KOH → NH₂SO₃K + H₂O. A)3.57 × 10^-4 N B)1.77 × 10^-4 N C)4.38 × 10³ N D)5.64 N E)0.177 N

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Quizplus · Accepted Answer

Normality (N) is defined as the number of equivalents of solute per liter of solution. For the reaction given, sulfamic acid (HNSO3H) reacts with KOH in a 1:1 molar ratio, meaning 1 mole of sulfamic acid reacts with 1 mole of KOH. First, calculate the moles of sulfamic acid:- The molar mass of sulfamic acid (HNSO3H) is approximately 97.09 g/mol.- Moles of sulfamic acid = mass / molar mass = 0.477 g / 97.09 g/mol ≈ 0.00491 mol.Since the reaction is 1:1, the moles of KOH needed for neutralization are also 0.00491 mol.To find the normality of the KOH solution:- Volume of KOH solution = 27.7 mL = 0.0277 L.- Normality (N) = moles of solute / liters of solution = 0.00491 mol / 0.0277 L ≈ 0.177 N.Therefore, the normality of the potassium hydroxide solution is approximately 0.177 N.

Calculate the Normality of a Potassium Hydroxide Solution If 27