Chemistry question

Question

Al + FeO ⇒ Al2O3 + Fe

Step one: balance equation:

Please answer the following in number only; example, 4

What is the coefficient for Al? 

What is the coefficient for FeO?                

What is the coefficient for Al2O3?   

What is the coefficient for Fe?   

Step two: find the molar mass

What is the molar mass of FeO?  

What is the molar mass of Al2O3?  

Step three: Calculation

How many grams of Al2O3 is extracted from 250. g of FeO?  

Answer 1

Balanced reaction: 2 Al + 3 FeO --> Al₂O₃ + 3 Fe, from which the coefficients are 2, 3, 1, 3.

Molar masses FeO = 71.84 g, Al₂O₃ = 101.96 g.

(250. g FeO) * (1 mole FeO / 71.84 g FeO) * (1 mole Al₂O₃ / 3 mole FeO) * (101.96 g Al₂O₃ / 1 mole Al₂O₃) = 118 g Al₂O₃.

Answer 2

Step 1:Let's balance O first: I see O₃ on the product side and only one O in compound FeO, so let's give FeO a coefficient of 3.

Next, we look at Fe. We now have 3 Fe on the left side from the previous step so let's also put a 3 coefficient on the product Fe to balance that.

Finally, we look at Al. We see 1 Al on the reactant and 2 on the product, so all we need to do is throw a 2 coefficient on the reactant Al to satisfy this.

Balancing should yield 2Al + 3FeO -> Al₂O₃ + 3Fe

Step 2: To find molar mass, we need each element's atomic mass:

O - 16.000 g/mol

Al - 26.982 g/mol

Fe - 55.845 g/mol

Now, sum these with subscripts to find molar masses.

FeO: 55.845 + 16 = 71.845 g/mol

Al₂O₃: 2(26.982) + 3(16) = 53.964 + 48 = 101.964 g/mol

Step 3: We know previously from balancing that it takes 3 moles of FeO to produce 1 mole of Al₂O₃. Keep this in mind while calculating.

First, let's find how many moles of FeO we have: 250 g / 71.845 g / mol = 3.48 moles of FeO.

Now, let's use our mole ratio. As noted at the beginning of this step, it is 3:1 so 3.48 moles of FeO produces 3.48 mol FeO * (1 mole Al₂O₃ / 3 mole FeO) = 3.48/3 = 1.16 moles of Al₂O₃ produced.

From here, all we do is multiply the molar mass of Al₂O₃ by the number of moles we have and we will arrive at our answer!

1.16 mol Al₂O₃ * (101.964 g / 1 mol) = 118 g Al₂O₃ produced (3 sig figs due to 250. in question)

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