What mass of salt is formed by the interaction of excess nitric acid and 60 g of aluminum oxide containing 5% impurities?

Given:
m s (Al2O3) = 60 g
w p (Al2O3) = 5%
To find:
m (Al (NO3) 3)
Decision:
Al2O3 + 6HNO3 = 2Al (NO3) 3 + 3H2O
w o.c. (Al2O3) = 100% -5% = 95% = 0.95
m in (Al2O3) = w p.w * m s = 0.95 * 60 g = 57 g
n (Al2O3) = m / M = 57 g / 102 g / mol = 0.56 mol
n (Al2O3): n (Al (NO3) 3) = 1: 2
n (Al (NO3) 3) = 0.56 mol * 2 = 1.12 mol
m (Al (NO3) 3) = n * M = 1.12 mol * 213 g / mol = 238.56 g
Answer: 238.56 g