a) Find the empirical formula for KxFey(C2O4)z·nH2O given % percents:
potassium = 34.5%
iron = 11.4%
water = 11.0%
oxalate = 43.1 %
b) Insert values for x, y, z into KxFey(C2O4)z·nH2O then give correct coefficients to balance the equation:
_Fe(NH4)2(SO4)2*6H20 + __H2C2O4 + __K2C2O4 + H2O2 --> __K3Fe(C2O4)3*3H20 + __(NH4)2SO4 + __H2SO4 + H20
I have this part, I know that it's
2 Fe(NH4)2(SO4)2*6H20 + 3 H2C2O4 + 3 K2C2O4 + 1 H2O2 --> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + 2 H2SO4 + 2 H20
c) Using the balanced equation and mass of Fe(NH4)2(SO4)2*6H20 (2.501g) determine the percent yield of KxFey(C2O4)z·nH2O.
To find the empirical formula for KxFey(C2O4)z·nH2O, we need to calculate the mole ratio of each element in the compound based on the percent composition.
a) Let's calculate the number of moles for each element:
1. Potassium (K):
Percent: 34.5%
Moles = (percent/100) * total mass (molecular weight)
Moles of K = (34.5/100) * total mass of K
2. Iron (Fe):
Percent: 11.4%
Moles of Fe = (11.4/100) * total mass of Fe
3. Oxygen (O):
Percent: present in oxalate (C2O4)
Moles of O = (43.1/100) * total mass of C2O4 * 4 (since there are 4 oxygen atoms per molecule of C2O4)
4. Hydrogen (H):
Percent: present in water (H2O)
Moles of H = (11.0/100) * total mass of H2O * 2 (since there are 2 hydrogen atoms per molecule of H2O)
5. Carbon (C):
The remaining percentage is the percent of carbon in the compound:
Percent of C = 100 - (Percent of K + Percent of Fe + Percent of O + Percent of H)
Moles of C = (Percent of C/100) * total mass of C2O4 * 2 (since there are 2 carbon atoms per molecule of C2O4)
Now we have the moles of each element, and we need to find the simplest whole number ratio. Divide each mole value by the smallest mole value to obtain the ratio.
The empirical formula for the compound will be KxFey(C2O4)z·nH2O in the determined ratio.
b) The balanced equation you provided is correct. Now we need to calculate the correct coefficients to balance the equation.
Fe(NH4)2(SO4)2 * 6H2O + 3H2C2O4 + 3K2C2O4 + H2O2 --> 2K3Fe(C2O4)3 * 3H2O + 2(NH4)2SO4 + 2H2SO4 + 2H2O
c) To find the percent yield of KxFey(C2O4)z·nH2O, we need the mass of the compound produced and the theoretical yield.
Using the balanced equation, we can see that the molar ratio of Fe(NH4)2(SO4)2 * 6H2O to K3Fe(C2O4)3 * 3H2O is 2:2, which means that for every 2.501g of Fe(NH4)2(SO4)2 * 6H2O, we should expect to produce 2.501g of K3Fe(C2O4)3 * 3H2O.
Assuming you have a certain mass of K3Fe(C2O4)3 · 3H2O produced, divide the actual mass by the theoretical mass and multiply by 100 to get the percent yield.
Percent Yield = (Actual mass / Theoretical mass) * 100
To find the empirical formula for KxFey(C2O4)z·nH2O, we need to determine the mole ratio between the elements. This can be done by converting the percentages to grams and calculating the number of moles.
a) First, convert the percentages to grams:
Potassium = 34.5% = 34.5 grams
Iron = 11.4% = 11.4 grams
Water = 11.0% = 11.0 grams
Oxalate = 43.1% = 43.1 grams
Next, calculate the moles of each element by dividing the mass by the molar mass:
Molar mass of potassium (K) = 39.10 g/mol
Molar mass of iron (Fe) = 55.85 g/mol
Molar mass of water (H2O) = 18.02 g/mol
Molar mass of oxalate (C2O4) = 88.02 g/mol
Moles of potassium (K) = 34.5 g / 39.10 g/mol = 0.882 mol
Moles of iron (Fe) = 11.4 g / 55.85 g/mol = 0.204 mol
Moles of water (H2O) = 11.0 g / 18.02 g/mol = 0.610 mol
Moles of oxalate (C2O4) = 43.1 g / 88.02 g/mol = 0.489 mol
Now, divide the moles by the smallest number of moles to get the mole ratio:
Mole ratio of potassium (K) = 0.882 mol / 0.204 mol = 4.32 (approximately)
Mole ratio of iron (Fe) = 0.204 mol / 0.204 mol = 1
Mole ratio of water (H2O) = 0.610 mol / 0.204 mol = 2.99 (approximately)
Mole ratio of oxalate (C2O4) = 0.489 mol / 0.204 mol = 2.39 (approximately)
Now, simplify the mole ratios by dividing them by the smallest value (in this case, 1):
Mole ratio of potassium (K) = 4.32 (approximately)
Mole ratio of iron (Fe) = 1
Mole ratio of water (H2O) = 2.99 (approximately)
Mole ratio of oxalate (C2O4) = 2.39 (approximately)
Therefore, the empirical formula for KxFey(C2O4)z·nH2O is approximately K4.32Fe(C2O4)2.39(H2O)2.99.
b) Based on the balanced equation you provided, the correct coefficients to balance the equation are as follows:
2 Fe(NH4)2(SO4)2*6H20 + 3 H2C2O4 + 3 K2C2O4 + 1 H2O2 → 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + 2 H2SO4 + 2 H20
c) To determine the percent yield of KxFey(C2O4)z·nH2O, you need to know the mass of the desired product formed. Unfortunately, the given mass refers to Fe(NH4)2(SO4)2*6H20, which is one of the reactants.
Without the mass of the desired product, it is not possible to calculate the percent yield. The percent yield is determined by dividing the actual yield (mass of desired product obtained in the experiment) by the theoretical yield (maximum possible yield based on stoichiometry) and multiplying by 100.
In order to calculate the percent yield, you need to conduct the experiment and measure the mass of the actual product obtained. Then, you can use the balanced equation to determine the theoretical yield and calculate the percent yield using the formula mentioned above.
bantuan membuka kata sandi
a) Find the empirical formula for KxFey(C2O4)z·nH2O given % percents:
potassium = 34.5%
iron = 11.4%
water = 11.0%
oxalate = 43.1 %
Take a 100 g sample. That will give you
34.5 g K
11.4 g Fe
11.0 g H2O
43.1 g oxalate (C2O4)
Now divide mols = g/molar mass.
From mols of each, determine the ratio of one to the other. The easy way to do this is to divide the smallest # mols by itself. That will make that element/polyatomic ion 1.000. Then, to keep things equal, divide the other mols by that same number, then round to the enarest whole number (but I caution not to round numbers from 0.4 to 0.6 up or down). That will give you the formula. Do that part, then we can talk about the remainder. You may be able to solve the rest of the problem after getting the formula.