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The reaction between aluminum and iron(iii) oxide can generate temperatures approaching 3000°c and is used in welding metals: 2al + fe2o3 → al2o3 +2fe in one process, 149 g of al are reacted with 601 g of fe2o3. calculate the mass (in grams) of al2o3 formed, and determine the amount of excess reagent left at the end of the reaction.

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First, we need to find the limiting reactant. 

149 g Al x 1 mole/27g Al = 5.52 mol Al 

601 g Fe2O3 x 1 mole/159.6g Fe2O3 = 3.77 mol Fe2O3 

Al is the limiting reactant since 3.77 mol Fe2O3 would require 3.77*2 = 7.54 mol Al but the given is less.

mol Al2O3 = 5.52 mol Al  * 1 mol Al2O3 / 2 mol Al = 2.76 mol

mass Al2O3 = 2.76 mol x 102g/1 mole AlO3 = 281.52 g Al2O3 will be formed. 



Since 3.77 mole of Fe2O3 is present, but 5.52 mole Al is the limiting reactant, then only 5.52/2 mole Fe2O3 can be used. This leaves an excess of:

Fe2O3 excess = 3.77 – 5.52/2 = 1.01 

1.01 mole Fe2O3 x 103.8g/1 mole Fe2O3 = 104.84 g Fe2O3 

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