Posted by Natash on .
Okay so Im doing a lab on preparation of K3[Fe(C2O4)3]*3H2O The final weight I got 2.835g. How can I find the theoretical yield of this? What do I need?
1.You need the balanced equation.
2.Convert grams of the starting material to moles. moles = grams/molar mass. Be sure to use as the starting material the one that was NOT in excess.
3.Using the coefficients in the balanced equation, convert moles of the starting material to moles of the final product.
4. Now convert moles of final product to grams. grams = moles x molar mass. That number of grams is the theoretical yield.
5. I assume you will want to find the percent yield. That will be
percent yield = (actual yield/theoretical yield)*100. The actual yield, of course, is 2.835 grams.
sorry I'm a bit confused. These are the solutions we had to prepare
a. 10g of ferrous ammonium sulfate dissolved in 30mL of water containing 2 drops of 6M H2SO4. Which I got 8.87g
b. 6g of oxalic acid in 50mL of water. I got 4.4008g
c. 6.6 g of K2C2O4*H2O in 18mL of water. I got 4.9142
d.1.7 oxalic acid in 15mL of water. I got .0471g
the balanced equation is Fe(H2O)4(C2O4)^-2 + 2(C2O4)2- +3K+ +3H2O---> K3[Fe(C2O4)3]*3H2O
is the starting material the oxalic acid ?
Frankly, I don't know what you are doing. The equation isn't balanced (the charges don't balance), I don' see the NH4^+ anywhere in the equation, it appears you have used the starting material with (H2O)4 and as an ion. I assume that is the ferrous ammonium sulfate and finally, the oxidation state of the Fe on the left is +2 an that on the right is +3 but I don't see anything reduced. Both solution b and d are oxalic acid. If you want to type in the procedure I will try to figure out what you are doing. Is the ferrous ammonium sulfate Fe(NH4)2(SO4)2?
yes Fe(NH4)2(SO4)2 is the ferrous ammonium sulfate.
here is the procedure, if you can try to figure it out that'll be great thank you!
1. prepare and label the following solutions:
a.10g of ferrous ammonium sulfate dissolved in 30mL of water containing 2 drops of 6M H2SO4
b. 6g of oxalic acid in 50mL of water
c.6.6g of K2C2O4*H2O in 18mL of water
d. 1.7g oxalic acid in 15mL of water
2.carry out the following preparation in a 250mL beaker. To the ferrous ammonium sulfate solution add, with stirring, the solution of 6 g acid in water. A yellow precipitate of iron(II) oxalate forms quickly. Carefully heat the mixture to near boilin, stirring constantly to avoid bumping. Allow the precipitate to settle, decant and discard the supernatant liquid.
3.Wash the precipitate 3x by adding 30mL of hot water, stirring and decanting the liquid.
4.Add the 6.6g potassium oxalate /18mL water solution to the wet iron(II) oxalate precipitate and heat the mixture to about 50 C. Slowly add 17mL of 6% H2O2 while stirring and maintaining a temp. of about 50 C for at least 15 mins or until all unconverted iron(II) oxalate is converted.
5.heat the mixture to boiling and slowly add the 1.7g oxalic acid in 15mL water solution; maintain the temperature near boiling for 10 mns. filter away any precipitate before adding ethanol. Add 20mL 95% ethanol to the filtrate. cover the beaker with a watchglass, wrap in aluminium foil and place the beaker in your locker until next week.
6. after 1 week, filter the green product crystals with a buchner funnel, and wash twice with a few mL of 50% aqueous ethanol, then with a few mL of 95% ethanol. continue to aspirate the crystals on the filter for 30 mins to air dry your samples. cover with aluminum foil during drying to prevent photcomposition. when dry, isolate and weigh your sample to determine the % yield.
Thank you soo much