Question

How many grams are of aluminum nitrate are contained in 0.150 moles of Al(NO3)3? ​

Answer 1

31.95 g

Step-by-step explanation:

find the molar mass of Al(NO3)3

(Al = 27, N = 14, O = 16)

molar mass of Al(NO3)3 = 27+ (14+16×3)×3

= 213 gmol^-

mass = 213 ×0.15

= 31.95g