Answer:
<u><em>This answer assumes that the strated "16.0g of iron" was meant to be 16.0 grams of iron(III) oxide.</em></u>
Explanation:
To start, the thermite equation must be balanced.
I find:
1Fe2O3 + 2Al = 1Al2O3 + 2Fe
This tells us we need 2 moles of Al for every 1 mole of Fe2O3.
Now calculate the moles of each reactant:
Moles Fe2O3: 16.0 g/159.7 g/mole = <u>0.100 moles Fe2O3</u>
Moles Al: 8.1 /26.98 g/mole = <u>0.300 moles Al</u>
The balanced equation says that in order to react all of the Fe2O3 we'd need twice that amount (in moles) of the Al. (0.100 moles Fe2O3)*(2) = 0.200 moles Al.
<u>Which of the two reactants is the limiting reagent?</u>
We have more than enough moles of Al to react with 0.10 moles of Fe2O3. (We have 0.300 moles Al and all we need is 0.200 moles to react with the 0.10 moles of Fe2O3. <em>Fe2O3 is the limiting reagent.</em>
<u><em>Calculate the maximum mass of iron of iron that could be formed using these quantities of reactants.</em></u>
The balanced equation tells us that we will obtain 2 moles of Fe for every 1 mole of Fe2O3 consumed. Since Fe2O3 is the limiting reagent, we will assume that it completely reacts. That means 0.1 moles of Fe2O3 is reacted. Since we expect twice that many moles of Fe, we should obtain 0.200 moles of Fe. At 55.85 g/mole, we should obtain:
(0.200 moles Fe)*(55.85 g Fe/mole Fe) = 11.2 grams Fe
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