percent yield, theoretical yield, and actual yield of iron

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Oct 23rd, 2015

Hi, first you want to find the actual yield you got of iron. The problem stated that you produced 15.8 g of Fe. That is your actual yield, you just need to convert it to moles. In order to you this you need to divide 15.8 g Fe by the molecular weight of Fe, which looking at a periodic table you can find to be 55.845 g/mol. So you do this and get 15.8 g Fe / 55.845 g/mol Fe = 0.283 mol (3 sig figs). This is your actual yield of iron.

The theoretical yield is the number of moles of product you would get if everything in the reaction went perfectly. the number of the limiting reactant in moles. The limiting reactant is the reactant you have the least amount of and is what will stop the reaction when it runs out. Since the problem states you have excess carbon that means that the 0.321 mol of Fe2O3 is the limiting reagent. This means that your theoretical yield is 0.321 mol.

The percent yield is the actual yield divided by the theoretical yield multiplied by 100. We know both of those values so if we plug them in we get:

(0.283 mol / 0.321 mol) * 100 = 88.2% (for 3 sig figs)

Oct 23rd, 2015

when i enter this in, the theoretical and percent yield are both not correct... and I'm confused as to why

Oct 23rd, 2015

when i enter this in, the theoretical and percent yield are both not correct... and I'm confused as to why

Oct 23rd, 2015

when i enter this in, the theoretical and percent yield are both not correct... and I'm confused as to why

Oct 23rd, 2015

Oops I think I forgot to factor in the molar ratios. One second.

Oct 23rd, 2015

Sorry about that. I think I know where I went wrong. So for theoretical yield you have to get the ratio of moles Fe2O3 to moles Fe. If you look at the numbers or coefficients in front of the reactants and products you see that there is a 1 in front of Fe2O3 (there is actually nothing there so that means there is a 1 in front). You can see that there is a 2 in front of Fe. We have to use these to form a ratio.

So you take the 0.321 moles Fe2O3 and do a ratio like this:

(0.321 moles Fe2O3) X (2 moles Fe / 1 mole Fe2O3)

This is considered stoichiometry and will give us the correct ratio. This gives us 0.642 moles Fe and is the limiting reactant.

So you now have a theoretical yield of 0.642 mol. If we plug that into the same % yield equation we have (0.283 mol / 0.642 mol) X 100 = 44.1 % (3 sig figs).

Try these values and let me know if they arent working or anything.

Oct 23rd, 2015

Did those values work?

Oct 23rd, 2015

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Oct 23rd, 2015
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Oct 23rd, 2015
Oct 21st, 2017
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