63.75 g of alumina containing 20% impurities was treated with an excess of nitric acid.

63.75 g of alumina containing 20% impurities was treated with an excess of nitric acid. calculating the mass of the salt being reversed.

1. Let’s get rid of impurities in aluminum oxide, making up the proportion:

63.75 g – 100%

x g – 80%

Hence, x = 63.75 * 80/100 = 51 g Al2O3

2. Let’s compose the reaction equation:

Al2O3 + 6 HNO3 = 2 Al (NO3) 3 + 3 H2O

According to the equation:

Al2O3 – 1 mol;
Al (NO3) 3 – 2 mol.
Let’s find the molar mass of Al2O3 by the formula:

m (Al2O3) = n * M = 1 * (27 * 2 + 16 * 3) = 102 g

Let’s find the molar mass of Al (NO3) 3 by the formula:

m (Al (NO3) 3) = n * M = 2 * (27 + (14 + 16 * 3) * 3) = 2 * 213 = 426 g

3. Let’s calculate the mass of salt, making up the proportion:

51 g Al2O3 – x g (Al (NO3) 3)

102g Al2O3 – 426g (Al (NO3) 3)

Hence, x = 51 * 426/102 = 213 g.