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Fe2O3(s) + 2 Al(s) 2 Fe(l) + Al2O3(s)
Suppose we want to make 18.0 grams of iron. We must figure out how much starting material we need. This is a three-step calculation. Let's go through these steps one-by-one to calculate the mass of Fe2O3 that we must start with.
Explanation of how to get this answer would be appreciated more than just an answer, thanks in advance.
Write the equation and balance it. That's done BUT you didn't put an arrow in. Unless I knew where it went, it wouldn't make sense. Remember an equation is not complete without an arrow to separate the reactants from the products.
Step 2. Convert 18 g Fe to moles. moles = g/molar mass.
Step 3. Using the coefficients in the balanced equation, convert moles of what you have (in this case Fe) to moles of what you want (in this case Fe2O3).
Step 4. Convert moles Fe2O3 to grams.
grams = moles x molar mass.
I had four steps because I counted the equation as one.
Check my thinking. Check my work. Post your work if you get stuck.
Please elaborate on step 3.
The balanced chemical equation is:
Fe2O3(s) + 2 Al(s) -> 2 Fe(l) + Al2O3(s)
After step 2 I have .32 moles Fe.
If you have 18 g (stated that way), then 0.32 mole Fe is ok but if it is 18.0 you should carry another place which will be 0.322 mole Fe.
Step 3. Using the coefficients in the balanced equation, convert moles of what yu have (Fe) into moles of what you want (Fe2O3). This is just dimensional analysis.
moles Fe x (1 mole Fe2O3/2 moles Fe) or
0.32 moles Fe x (1 mole Fe2O3/2 moles Fe) = 0.32 x (1/2) = 0.16 mole Fe2O3.
Note how the factor is placed so that moles Fe in the denominator of the factor cancel moles Fe in the numerator of 0.32 moles Fe which leaves mole Fe2O3. The coefficients in the equation are the multipliers in the factor of 1/2. Note that we had moles Fe, we wanted to convert to moles Fe2O3 so we arranged the factor so that moles Fe canceled and the only unit left standing is moles Fe2O3. All conversions from one part of a chemical equation to any other is done just this way. For example, we could have converted moles Fe to moles Al.
0.32 mole Fe x (2 moles Al/2 moles Fe) = 0.32 x 1/1 = 0.32 moles Al were used in the reaction OR
how much Al2O3 was produced?
0.32 moles Fe x (1 mole Al2O3/2 moles Fe) = 0.32 x 1/2 = 0.16 mole Al2O3.
Got it now, thanks a ton.