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Answer to Question #65335 in Chemistry for preston
Question #65335
What mass of oxygen (931.998 g/mol) would be needed to react with exactly 0.0154 mol of aluminum (26.982 g/mol)?
Expert's answer
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
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
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