Calculate the mass of nitric acid (HNO3) required to neutralize aluminum hydroxide (ALOH) with a mass of 0.7 g

Neutralization reactions:

3HNO3 + Al (OH) 3 = Al (NO3) 3 + 3H2O

The mass of aluminum hydroxide is 0.7 g. Let us find the amount of mol of aluminum hydroxide by dividing the mass of aluminum hydroxide by its molar mass.

M (Al (OH) 3) = 78 g / mol.

n (Al (OH) 3) = m (Al (OH) 3) / M (Al (OH) 3) = 0.7 / 78 = 0.009 mol.

According to the reaction equation, 3 mol of HNO3 reacts with 1 mole of Al (OH) 3. This means that 0.009 * 3 = 0.027 mol of HNO3 interacts with 0.009 mol of Al (OH) 3.

Let’s find the mass of nitric acid.

M (HNO3) = 63 g / mol.

m (HNO3) = n (HNO3) * M (HNO3) = 0.027 * 63 = 1.7 g.