Question

How many grams of aluminum Al are needed to produce 12.0 G of aluminum oxide (A12 O3)​

Answer 1

4Al + 3O₂ ⇒ 2Al₂O₃

n(Al₂O₃) = m/M = 12/101.96 = 0.117693213mol

unknown/known = 2/4 = 1/2

n(Al) = n(Al₂O₃)/2 = 0.05884660651mol

m(Al) = n x M = 0.05884660651 x 26.98 = 1.59g (2dp)