Calculate the mass of aluminum required for the reduction of chromium from 304 g of chromium oxide (lll)
| April 19, 2021 | education
Calculate the mass of aluminum required for the reduction of chromium from 304 g of chromium oxide (lll) by the method of aluminothermy (reduction with aluminum).
Decision:
Cr2O3 + 2Al = Al2O3 + 2Cr
v (Cr2O3) = 304g / 152g mol = 2 mol
v (Al) = 2 * v (Cr2O3) = 2 mol * 2 = 4 mol
m (Al) = 4 mol * 27 g mol = 108 g
Answer: m (Al) = 108 g
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