What mass of solid magnesium hydroxide is needed to neutralixe .385 Liters of 2.5 M nitric acid?
To solve this problem, we need to use the balanced chemical equation for the reaction between nitric acid (HNO3) and magnesium hydroxide (Mg(OH)2):
2 HNO3 + Mg(OH)2 -> Mg(NO3)2 + 2 H2O
From the equation, we can see that 2 moles of nitric acid react with 1 mole of magnesium hydroxide.
Step 1: Calculate the number of moles of nitric acid.
The given concentration of nitric acid is 2.5 M, which means there are 2.5 moles of nitric acid in 1 liter of the solution. Since we have 0.385 liters of the solution, we can calculate the number of moles of nitric acid as follows:
0.385 L x 2.5 moles/L = 0.9625 moles of nitric acid
Step 2: Determine the amount of magnesium hydroxide needed.
From the balanced chemical equation, we know that 2 moles of nitric acid react with 1 mole of magnesium hydroxide. Therefore, the number of moles of magnesium hydroxide needed will be half the number of moles of nitric acid:
0.9625 moles of nitric acid / 2 = 0.48125 moles of magnesium hydroxide
Step 3: Calculate the mass of magnesium hydroxide.
To calculate the mass of magnesium hydroxide, we need to know its molar mass. The molar mass of magnesium hydroxide can be calculated by adding up the atomic masses of its constituent elements:
Mg: 24.31 g/mol
O: 16.00 g/mol (x 2)
H: 1.01 g/mol (x 2)
Molar mass of Mg(OH)2 = (24.31 g/mol) + (16.00 g/mol x 2) + (1.01 g/mol x 2) = 58.33 g/mol
Finally, we can calculate the mass of magnesium hydroxide needed using its molar mass and the number of moles calculated earlier:
Mass of Mg(OH)2 = 0.48125 moles x 58.33 g/mol ≈ 28.08 grams
Therefore, approximately 28.08 grams of solid magnesium hydroxide is needed to neutralize 0.385 liters of 2.5 M nitric acid.