Calculate the mass of nitric acid required to neutralize 15.6 g of aluminum hydroxide.

Let’s implement the solution:
1. According to the condition of the problem, we compose the equation:
m = 15.6 g. X g. -?
Al (OH) 3 + 3HNO3 = Al (NO3) 3 + 3H2O – ion exchange, aluminum nitrate, water is released;
2. Calculation of molar masses:
M Al (OH) 3 = 77.9 g / mol;
M (HNO3) = 63 g / mol.
3. Determine the amount of the original substance by the formula:
Y Al (OH) 3 = m / M = 16.6 / 77.9 = 0.2 mol.
4. Proportion:
0.2 mol Al (OH) 3 – X mol (HNO3);
-1 mol – 3 mol from here, X mol (HNO3) = 0.2 * 3/1 = 0.6 mol.
5. Find the mass of acid:
m (HNO3) = Y * M = 0.6 * 63 = 37.8 g.
Answer: to carry out the process, nitric acid with a mass of 37.8 g is required.