Find the mass of salt in the reaction between aluminum hydroxide and 79 g of nitric acid.

We carry out the solution:

By the condition of the problem, we write down the process:
Al (OH) 3 + 3HNO3 = Al (NO3) 3 + 3H2O – ion exchange, aluminum nitrate was formed;

Let’s make calculations using the formulas:
M (HNO3) = 63 g / mol;

M Al (NO3) 3 = 150.9 g / mol;

Y (HNO3) = m / M = 1.25 mol.

Proportion:
1.25 mol (HNO3) – X mol Al (NO3) 3;

-3 mol                      -1 mol hence, X mol of Al (NO3) 3 = 1.25 * 1/3 = 0.42 mol.

Find the mass of the product:
m Al (NO3) 3 = Y * M = 0.42 * 150.9 = 63.4 g

Answer: received aluminum phosphate weighing 63.4 g