Question

A new base has been discovered, narium hydroxide: Nr(OH)₃ how many grams of this base are needed to neutralize 96.5 ml of 0.556 m H₂SO₄(aq)? *do not list units in your answer* molar mass Nr(OH)₃ = 77.99 g/mol

Answer 1

To neutralize 0.556 M of H₂SO₄ using narium hydroxide (Nr(OH)₃), you need 6.277 grams of Nr(OH)₃, accounting for the 3:2 mole ratio between Nr(OH)₃ and H₂SO₄.

Step-by-step explanation:

To calculate the mass of narium hydroxide (Nr(OH)₃) needed to neutralize sulfuric acid (H₂SO₄), we need to consider the stoichiometry of the reaction. According to the balanced chemical equation, it takes three moles of hydroxide (OH-) to neutralize two moles of H₂SO₄, thus the reaction is:

2 H₂SO₄ (aq) + 3 Nr(OH)₃ (aq) → Nr₃(SO₄)₂ (aq) + 6 H₂O (l)

We first calculate the number of moles of H₂SO₄ using its concentration and volume:

Moles H₂SO₄ = 0.556 mol/L × 0.0965 L = 0.053654 moles H₂SO₄

Since the molar ratio of Nr(OH)₃ to H₂SO₄ is 3:2, we need 1.5 times more moles of Nr(OH)₃:

Moles Nr(OH)₃ = 1.5 × Moles H₂SO₄ = 1.5 × 0.053654 moles = 0.080481 moles Nr(OH)₃

Finally, we calculate the mass of Nr(OH)₃ needed:

Mass Nr(OH)₃ = 0.080481 moles × 77.99 g/mol = 6.277 g