This is a balancing eqn Im finding rather tricky:
___Fe(NH4)2(SO4)2 6H2O + ___ H2C2O4 + ___K2C2O4 + ___ H2O2 ---> ___K5Fe1(C2O2)2 3H2O + ___(NH4)2(SO4) + ___H2SO4 + ___ H2O
remember H2O2, an oxidizing agent, was used in synthesis, so youll nee to do a redox balance. Start with just iron ions, balance the rxn, then add the full formulas.
Thanks!
my mistake, the first molecule in the products should be K3Fe(C2O4)3 3H20
Check this out. You fill in the blank for water.
2,3,3,1,==>2,2,2,??H2O
I have been having trouble with this as well. Would the coefficient of water be 5,8, or 2? I have heard all of these and I got 5 when I worked it out, but I have been told that it is wrong.
To balance this equation, we need to follow these steps:
1. Start by balancing the atoms that appear in only one compound on each side of the equation.
2. Balance the atoms that appear in more than one compound. Begin with the metals, then balance the nonmetal atoms.
3. Finally, balance the charges by adding electrons to the side that has a higher positive charge.
Let's begin balancing the equation:
First, let's count the atoms on both sides of the equation:
Left side: 1 Fe, 2 nitrogen (N), 6 hydrogen (H), 8 oxygen (O), 2 sulfur (S), 2 carbon (C), 5 potassium (K), and 28 hydrogen (H).
Right side: 5 potassium (K), 1 iron (Fe), 4 carbon (C), 4 oxygen (O), 2 nitrogen (N), 3 hydrogen (H), 1 sulfur (S), and 30 hydrogen (H).
Now we can follow the steps to balance the equation:
1. Start by balancing the atoms that appear in only one compound on each side of the equation.
There are 2 Fe atoms on the left side, but only 1 Fe atom on the right side. To balance this, we need to put a coefficient of 2 in front of K5Fe1(C2O2)2 3H2O on the right side.
Now our equation looks like this:
___Fe(NH4)2(SO4)2 6H2O + ___ H2C2O4 + ___K2C2O4 + ___ H2O2 ---> 2K5Fe1(C2O2)2 3H2O + ___(NH4)2(SO4) + ___H2SO4 + ___ H2O
2. Balance the atoms that appear in more than one compound.
- To balance the nitrogen (N) atoms, we need to put a coefficient of 3 in front of (NH4)2(SO4) on the right side.
- To balance the oxygen (O) atoms, we need to put a coefficient of 9 in front of H2O2 on the left side.
- To balance the potassium (K) atoms, we need to put a coefficient of 3 in front of K2C2O4 on the left side.
- To balance the sulfur (S) atoms, we need to put a coefficient of 3 in front of H2SO4 on the right side.
The equation now becomes:
___Fe(NH4)2(SO4)2 6H2O + ___ H2C2O4 + 3K2C2O4 + 9 H2O2 ---> 2K5Fe1(C2O2)2 3H2O + 3(NH4)2(SO4) + 3H2SO4 + ___ H2O
3. Balance the charges by adding electrons to the side that has a higher positive charge.
Since H2O2 is an oxidizing agent, it causes electrons (e-) to be transferred. In this case, it causes the iron (Fe) in (NH4)2(SO4)2 to undergo oxidation. To balance the charges, we need to add 10 electrons (10e-) to the left side.
The final balanced equation is:
___Fe(NH4)2(SO4)2 6H2O + ___ H2C2O4 + 3K2C2O4 + 9 H2O2 + 10e- ---> 2K5Fe1(C2O2)2 3H2O + 3(NH4)2(SO4) + 3H2SO4 + ___ H2O
Since we started with just iron ions, we will now substitute the full formula for Fe(NH4)2(SO4)2 as (NH4)2Fe(SO4)2.
The final balanced equation with full formulas is:
(NH4)2Fe(SO4)2 6H2O + H2C2O4 + 3K2C2O4 + 9 H2O2 + 10e- ---> 2K5Fe1(C2O2)2 3H2O + 3(NH4)2(SO4) + 3H2SO4 + ___ H2O
Note: The coefficients for H2C2O4, K2C2O4, and H2O can be discerned by comparing their respective atoms' coefficients.