27 g of aluminum was burned in 27 g of oxygen. What substance was taken in excess?
| December 3, 2020 | education
Let’s write the reaction equation: 4Al + 3O2 = 2Al2O3 n (Al) = m \ M = 27 \ 27 = 1 mol n (O2) = m \ M = 27 \ 32 = 0.84 mol To determine which substance is in excess, you need to divide the quantities substances by the coefficients in front of them in the reaction equation: 1: 4 = 0.25 0.84 \ 3 = 0.28 0.28> 0.25 Therefore, oxygen was taken in excess
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