4.0 g of ferrous ammonium sulphate, FeS04(NH4)2 SO4 6H20, is used. The oxalate is in excess, calculate the theoretical yield of the iron complex. Can someone please show me how to calculate a question like this please?
Thank you
1. Write the equation and balance it.
2. Convert 4.0 g of the ferrous ammonium sulfate to moles. #moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles of the ferrous ammonium sulfate to moles of the iron complex.
4. Finally, convert moles of the iron complex to grams. # grams = moles x molar mass. The number of grams of the iron complex is the theoretical yield.
How many moles of ammonium sulfate were released in 25.3 kg?
To calculate the theoretical yield of the iron complex, we need to use a stoichiometric calculation based on the balanced chemical equation for the reaction between ferrous ammonium sulphate and oxalate.
The balanced chemical equation for the reaction is:
FeSO4(NH4)2SO4·6H2O + K2C2O4 + H2SO4 → FeC2O4 + (NH4)2SO4 + K2SO4 + 6H2O
From the balanced equation, we can see that 1 mole of ferrous ammonium sulfate reacts with 1 mole of oxalate to produce 1 mole of the iron complex (ferrous oxalate).
Step 1: Convert the given mass of ferrous ammonium sulfate to moles.
To calculate the moles, we can use the molar mass of ferrous ammonium sulfate. The molar masses of the elements are:
Fe: 55.845 g/mol
S: 32.07 g/mol
H: 1.008 g/mol
O: 16.00 g/mol
N: 14.01 g/mol
The molar mass of ferrous ammonium sulfate is:
Fe: 1 * 55.845= 55.845 g/mol
S: 1 * 32.07= 32.07 g/mol
O: 4 * 16.00= 64.00 g/mol
N: 2 * 14.01= 28.02 g/mol
H: 20* 1.008= 20.16 g/mol
Adding these up gives a molar mass of:
55.845 g/mol + 32.07 g/mol + 64.00 g/mol + 28.02 g/mol + 20.16 g/mol= 200.10 g/mol
Now, we can calculate the number of moles:
moles = mass/molar mass
moles = 4.0 g / 200.10 g/mol
moles ≈ 0.01999 mol (rounding to five significant figures)
Step 2: Calculate the theoretical yield of the iron complex.
Since the reaction is 1:1 between ferrous ammonium sulfate and oxalate, the number of moles of the iron complex will be the same as the number of moles of ferrous ammonium sulfate.
So, the theoretical yield of the iron complex is approximately:
0.01999 mol (of iron complex) ≈ 0.01999 mol
Note: Remember that this is the theoretical yield, meaning it assumes 100% efficiency and complete conversion of the reactant. In actual practice, the yield might be lower due to side reactions or incomplete conversion.
To calculate the theoretical yield of the iron complex, you need to determine the stoichiometric ratio between the reactants and the product. In this case, the balanced chemical equation for the reaction between ferrous ammonium sulfate [(NH4)2Fe(SO4)2] and oxalic acid (H2C2O4) is:
(NH4)2Fe(SO4)2 + H2C2O4 → FeC2O4 + 2(NH4)2SO4 + H2O
From the equation, you can see that for every 1 mole of ferrous ammonium sulfate, you get 1 mole of iron oxalate.
Next, you need to convert the given mass of ferrous ammonium sulfate (4.0 g) to moles. To do this, you use the molar mass of ferrous ammonium sulfate:
Molar mass of (NH4)2Fe(SO4)2 = (2 x 18.02 g/mol from ammonium) + (55.85 g/mol from iron) + (2 x 32.07 g/mol from sulfur) + (8 x 16.00 g/mol from oxygen)
= 392.2 g/mol
Using this molar mass, you can calculate the number of moles of (NH4)2Fe(SO4)2:
Number of moles = mass / molar mass
Number of moles = 4.0 g / 392.2 g/mol
Number of moles ≈ 0.0102 mol
Since the stoichiometric ratio between (NH4)2Fe(SO4)2 and FeC2O4 is 1:1, the theoretical yield of iron complex (FeC2O4) is also 0.0102 mol.
To determine the mass of the theoretical yield, you can use the molar mass of iron oxalate (FeC2O4):
Molar mass of FeC2O4 = (55.85 g/mol from iron) + (2 x 12.01 g/mol from carbon) + (4 x 16.00 g/mol from oxygen)
= 143.88 g/mol
Mass of theoretical yield = number of moles × molar mass
Mass of theoretical yield = 0.0102 mol × 143.88 g/mol
Mass of theoretical yield ≈ 1.47 g
Therefore, the theoretical yield of the iron complex is approximately 1.47 grams when 4.0 grams of ferrous ammonium sulfate is used.