How much nitrate acid will react with 0.2 mol of aluminum hydroxide?

To solve the problem, we write the equation:
1. Al (OH) 3 + 3HNO3 = Al (NO3) 3 + 3H2O – ion exchange reaction, aluminum nitrate salt and water are released;
2. The ratio of substances in moles according to the equation are: 1 mol – aluminum hydroxide, 3 mol – nitric acid.
3. Let’s calculate the molar masses of substances:
M Al (OH) 3 = 77.9 g / mol;
M (HNO3) = 63 g / mol.
4. Let’s make a proportion according to the equation data and the problem conditions:
X mol (HNO3) – 0.2 mol Al (OH) 3;
-3 mol -1 mol hence, X mol (HNO3) = 3 * 0.2 / 1 = 0.6 mol.
Answer: the amount of nitric acid is 0.6 mol.