How much oxygen is required for the combustion of 200 g of technical aluminum (impurity content 2%)

univerkov | November 25, 2020 | education

M (al) tech = 200g
m (al) clean = 200 * (1-0.02) = 196 (g)
4al + 3o2 = 2al2o3
n (Al) = 196/27 = 7.26 (mol)
n (O2) = 5.445mol
V (O2) = 5.455 * 22.4 = 121.968 (L)

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