The reaction between aluminum and iron(iii) oxide can generate temperatures approaching 3000Â°c and is used in welding metals: 2al + fe2o3 â al2o3 +2fe in one process, 118 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. mass of al2o3 formed:

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barnuts barnuts · Answer 1 · 2017-10-21T06:39:30+02:00

We are given that balanced chemical reaction:

2 Al + Fe2O3 ---> Al2O3 + 2 Fe

1. To solve this, we must first determine the limiting reactant.

Calculate the number of moles of each reactant. (Al = 27 g/mol; Fe2O3 = 159.69 g/mol)

moles Al = 118 g / (27 g/mol) = 4.37 moles

moles Fe2O3 = 601 g / (159.69 g/mol) = 3.76 moles

The limiting reactant is the one which has lower moles/stoichiometric coefficient ratio:

Al = 4.37 / 2= 2.185

Fe2O3 = 3.76 / 1 = 3.76

So we can see that Al is the limiting reactant.

2. Calculate the mass (in grams) of Al2O3 formed

We based our calculation on the limiting reactant which is Al. From stoichiometry, 1 mole Al2O3 is formed for every 2 moles Al, therefore:

moles Al2O3 = 4.37 moles Al * (1 mole Al2O3 / 2 moles Al) = 2.185 moles Al2O3

Molar mass of Al2O3 is 101.96 g/mol, so mass is:

mass Al2O3 = 2.185 mol * 101.96 g/mol

mass Al2O3 = 222. 78 grams

3. The excess reagent is Fe2O3. From stoichiometry, 1 mole of Fe2O3 is consumed per 2 moles Al consumed, therefore:

moles Fe2O3 consumed = 4.37 moles Al * (1 mole Fe2O3 / 2 moles Al) = 2.185 moles

So remaining Fe2O3 is:

moles Fe2O3 remaining = 3.76 – 2.185 = 1.575 moles

In mass:

mass Fe2O3 remaining = 1.575 moles * (159.69 g/mol)

mass Fe2O3 remaining = 251.51 grams --> excess reactant