Question

What mass of oxygen (931.998 g/mol) would be needed to react with exactly 0.0154 mol of aluminum (26.982 g/mol)?

Answer 1

4Al +3O 2 = 2Al 2 O 3

n = m/M m = n·M

n (O 2 ) = ¾ n (Al) = ¾ 0.0154 = 0.012 mol

m (O 2 ) = 0.012 · 31.998 = 0.384 g