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13 Dec 2019

I did a chem lab where we made a coordination compound of Fe(III)KTOF3.

I need to figure out which is the limiting reagent, FeCl3, or KTOF3.

Here is the balanced equation: 3K2C204*H20+FeCl3. I know this means the equation has a 3:1 ratio.

I used 12.012 g of K2C204*H20, I know the molar mass is 184.231 g/mol. Which means I used 0.065201 moles of KTOF3 12.012g/ molar mass 184.231 is 0.065201 moles of KTOF3.

What is getting me stuck is the FeCl3. We used 7.9 ml of it, the concentration of it was 1.85 molar, and I know the molecular weight (126.751g).

How do I calculate the moles of FeCl3, so I can calculate the limiting reagent, theoretical yield, and the experimental yeild?

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I know the limiting reagent is the copper sulfate but I dont know how to find the theoretical yield and the theoretical mass of product produced. The molar mass is 245.75 g/mol) and I need it in moles.

1. The chemical equation for the synthesis, balanced only for the reactants, is as follows. Supply the coefficients for KCl and H2O required to balance the equation completely.

FeCl3·6H2O(aq) + 3K2C2O4·H2O(aq)

® K3Fe(C2O4)3·3H2O(s) + ____KCl(aq) + ____H2O(l)

Note that the reactants are used as their hydrates. FeCl3·6H2O means that one mol of the compound corresponds to one mol of FeCl3 and six mol of water. K2C2O4·H2O means that three mol of the compound corresponds to three mol K2C2O4 and three mol of water.

2. Calculate the amount (mols) corresponding to 5.00 g of FeCl3·6H2O (270.3 g/mol) and to 11.00 g of K2C2O4·H2O (184.2 g/mol). Note that the molar masses necessarily include the water of hydration.

3. On the basis of the chemical amounts calculated in no. 2 and the balanced chemical equation, determine the limiting reactant showing your calculations and telling why you chose that reactant.

4. Calculate the theoretical yield of the coordination compound (molar mass = 491.3 g/mol).

The product, like the reactants, is a hydrate. So the molar mass given is for the hydrate.

5. Calculate the percent yield based on an (assumed) actual yield of 7.52 g.


Reduction of 9-fluorenone: objective of this lab was to reduce 9-florenone(a ketone), to 9-fluorenol (an alcohol) using sodium borohydride as the reducing agent.

Experimental data:

9-fluorenone: MW 180.20 g/mol mass used 0.6 grams

methanol: MW 32.04 g/mol volume used: 6ml density: 0.792

NaBH4: MW 37.83 g/mol mass used 0.05 grams

9-fluorenol: MW 182.22 g/mol

Balanced equation for the reaction is: 4C13H8O + NaBH4+ 4CH3OH --> 4C13H10O + NaB(OCH3)4

Calculate the potential yield (in grams) if 9-fluorenol for 9-fluorenone, methanol and NaBH4. From these potential yields identify the limiting reagent and theoretical yield.

I tried calculating this on my own and I feel like I did this wrong. Please help

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