Calculate what mass of silver nitrate reacted with aluminum chloride if 4.26 g of aluminum nitrate was formed.

Given:
m (Al (NO3) 3) = 4.26 g
m (AgNO3) -?
Decision:
3AgNO3 + AlCl3 = Al (NO3) 3 + 3AgCl
M (Al (NO3) 3) = 27 + 42 + 144 = 213 g / mol
n = m / M, n (Al (NO3) 3) = 4.26 / 213 = 0.02 mol
3 mol Ag (NO3) 3 – 1 mol Al (NO3) 3
x mol – 0.02 mol
n (Ag (NO3) 3) = 0.02 * 3 = 0.06
M (Ag (NO3) 3) = 108 + 14 + 46 = 170 g / mol
m = n * M, m (AgNO3) 3) = 170 * 0.06 = 10.2 g
Answer: 10.2 g