The first step of the synthesis is described by the reaction below. When 2.000 g of Fe(NH4)2(SO4)2 6H2O is mixed with 13 mL of 1.0 M H2C2O4, the theoretical yield of FeC2O42H2O is ______
grams.
Fe(NH4)2(SO4)26H2O(s) + H2C2O4(aq)-->
FeC2O42H2O(s) + (NH4)2SO4(aq) + H2SO4(aq) + 4 H2O(l)
I didn't check it but make sure the equation is balanced.
Convert 2.000 g of the ferrous ammonium sulfate (the starting reactant) to mols remembering that mols = grams/molar mass.
Convert mols ferrous ammonium sulfate to mols of the product using the coefficients in the balanced equation.
Convert mols of the product to grams of the produce. mols x molar mass = grams.
That number of grams will be the theoretical yield of the product.
Post your work if you get stuck.
I've got the first part (0.0005), but I don't understand how to do the next 2 parts, where I covert mols ferrous ammonium sulfate to mols of the product. Thanks
I think you need to redo the math. I divided 2.000 g/392.14 = 5.1 x 10^-3. Make sure to confirm that. Perhaps I looked up the molar mass incorrectly.
Next, I didn't read the problem carefully enough for you must next determine the mols of H2C2O4 added. To do that, mols = M x L = 1.0 M x 0.013 L = 1.3 x 10^-2.
This is a limiting reagent problem Do you know how to do those?
I'll start you.
a)Use the coefficients in the balanced equation to start with EACH starting material to determine how many mols of the product can form. The first one is
0.0051 mole Fe(NH4)2(SO4)2.6H2O x (1 mol FeC2O4.2H2O/1 mol Fe(NH4)2(SO4)2.6H2O )= 0.0051 x 1/1 = 0.0051 mole FeC2O4.2H2O.
b)Next, do the same thing with mols H2C2O4.
0.013 mole H2C2O4 x (1 mol FeC2O4.2H2O/1 mol H2C2O4) = 0.013 x 1/1 = 0.013 mole FeC2O4.2H2O.
Note that the factor in each case is chosen to cancel the unit we start with and change it to the unit we want to keep. That is, in the first one note that Fe(NH4)2(SO4)2.6H2O is in the numerator AND denominator and they cancel. The numerator that is left is FeC2O4.2H2O and that is the unit we want at the end. In the case of the H2C2O4, note that mole H2C2O4 cancel and we keep the FeC2O4.2H2O in the numerator. The coefficients we use are the coefficients in the balanced equation.
c)Now compare the answer from a) and b) to see which is the smaller. That is the limiting reagent. In this case the 0.0051 mole FeC2O4.2H2O is the smaller number; therefore, that will be the amount of FeC2O4.2H2O formed and there will be some H2C2O4 that remains unreacted.
Step 4. Now convert mols to grams (g = mols x molar mass.
Step 5. That is the theoretical yield. Check my work.
To determine the theoretical yield of FeC2O42H2O, we need to first calculate the number of moles of Fe(NH4)2(SO4)2 6H2O and H2C2O4 involved in the reaction.
1. Calculate the number of moles of Fe(NH4)2(SO4)2 6H2O:
Given mass = 2.000 g
Molar mass of Fe(NH4)2(SO4)2 6H2O = (Fe: 55.845 g/mol) + 2(N: 14.007 g/mol) + 8(H: 1.008 g/mol) + 4(S: 32.06 g/mol) + 24(O: 16.00 g/mol) + 12(H2O: 18.015 g/mol) = 392.142 g/mol
Moles = Mass / Molar mass = 2.000 g / 392.142 g/mol
2. Calculate the number of moles of H2C2O4:
Given volume = 13 mL = 0.013 L
Molarity (M) = 1.0 M
Moles = Molarity x Volume = 1.0 M x 0.013 L
Now, using the balanced chemical equation, we can determine the stoichiometric ratio between Fe(NH4)2(SO4)2 6H2O and FeC2O42H2O. From the equation, we see that 1 mole of Fe(NH4)2(SO4)2 6H2O produces 1 mole of FeC2O42H2O.
So, if we have X moles of Fe(NH4)2(SO4)2 6H2O, the theoretical yield of FeC2O42H2O would be X moles as well.
3. From step 1, we calculated the number of moles of Fe(NH4)2(SO4)2 6H2O.
Therefore, the theoretical yield of FeC2O42H2O is also X moles.
Finally, if we want to convert moles to grams, we use the molar mass of FeC2O42H2O:
Molar mass of FeC2O42H2O = (Fe: 55.845 g/mol) + 2(C: 12.011 g/mol) + 4(O: 16.00 g/mol) + 8(H2O: 18.015 g/mol) = 244.158 g/mol
4. Calculate the theoretical yield of FeC2O42H2O in grams:
Theoretical yield (grams) = X moles x Molar mass
Note: The value of X is obtained in step 1.
By following these steps, you can calculate the theoretical yield of FeC2O42H2O.