how to balance this partial equation given below
k2cr2o7+h2so4+kI-->k2so4+cr2(so4)3+h2o+I2
To balance the partial equation:
1. Start by balancing the atoms that occur only once on either side of the equation.
K2Cr2O7 + H2SO4 + KI → K2SO4 + Cr2(SO4)3 + H2O + I2
Let's balance the sulfur (S) atoms first.
K2Cr2O7 + H2SO4 + KI → K2SO4 + Cr2(SO4)3 + H2O + I2
The reactant (H2SO4) has one sulfur atom, and the products have three sulfur atoms in the compound Cr2(SO4)3. To balance the sulfur atoms, we need to multiply the reactant (H2SO4) by 3:
K2Cr2O7 + 3H2SO4 + KI → K2SO4 + Cr2(SO4)3 + H2O + I2
2. Next, balance the oxygen (O) atoms. Start by counting the number of oxygen atoms on each side of the equation.
Reactant: 7 oxygen atoms (7 from K2Cr2O7)
Products: 16 oxygen atoms
- (4 from K2SO4 and 12 from Cr2(SO4)3)
To balance the oxygen atoms, we add water (H2O) to the reactant side of the equation. In this case, we need to add 9 water molecules.
K2Cr2O7 + 3H2SO4 + KI → K2SO4 + Cr2(SO4)3 + 9H2O + I2
3. Now, let's balance the hydrogen atoms. Start by counting the hydrogen atoms on each side of the equation.
Reactant: 10 hydrogen atoms (10 from H2SO4)
Products: 20 hydrogen atoms
- (4 from 2K2SO4 and 16 from 8H2O)
To balance the hydrogen atoms, we add hydrogen ions (H+) to the product side of the equation. In this case, we need to add 10 hydrogen ions.
K2Cr2O7 + 3H2SO4 + KI → K2SO4 + Cr2(SO4)3 + 9H2O + I2 + 10H+
4. Finally, balance any remaining atoms. In this case, we need to balance the potassium (K) atoms.
Reactant: 4 potassium atoms (4 from 2K2Cr2O7)
Products: 2 potassium atoms (2 from K2SO4)
To balance the potassium atoms, we add 2 more potassium ions (K+) to the product side of the equation.
K2Cr2O7 + 3H2SO4 + 2KI → K2SO4 + Cr2(SO4)3 + 9H2O + I2 + 10H+ + 2K+
Now the equation is balanced:
2K2Cr2O7 + 3H2SO4 + 2KI → K2SO4 + Cr2(SO4)3 + 9H2O + I2 + 10H+ + 2K+
To balance the given partial equation:
Step 1: Start by counting the number of atoms of each element on both sides of the equation.
On the left side:
K: 2 atoms
Cr: 2 atoms
O: 7 atoms
H: 2 atoms
S: 1 Atoms
I: 1 atom
On the right side:
K: 2 atoms
Cr: 2 atoms
O: 16 atoms (2 in K2SO4 and 14 in Cr2(SO4)3)
H: 2 atoms
S: 4 atoms
I: 2 atoms
Step 2: Balance the least abundant elements first. In this case, we can start with Iodine (I).
There are 2 iodine atoms on the left side, and only 1 iodine atom on the right side. To balance the iodine, we need to add a coefficient of 2 in front of KI.
k2cr2o7 + h2so4 + 2kI --> k2so4 + cr2(so4)3 + h2o + I2
Now the equation becomes:
K2Cr2O7 + H2SO4 + 2KI --> K2SO4 + Cr2(SO4)3 + H2O + I2
Step 3: Proceed to balance the next least abundant element, which is Oxygen (O).
There are 7 oxygen atoms on the left side, and 20 oxygen atoms on the right side. To balance the oxygen, we need to add a coefficient of 10/7 (approximately 1.43) in front of H2O.
k2cr2o7 + h2so4 + 2ki --> k2so4 + cr2(so4)3 + (10/7)h2o + I2
Now the equation becomes:
K2Cr2O7 + H2SO4 + 2KI --> K2SO4 + Cr2(SO4)3 + (10/7)H2O + I2
Step 4: Review the equation to check if any elements are still unbalanced.
All the elements are now balanced except for Hydrogen (H) and Sulfur (S).
On the left side, there are 2 hydrogen atoms, while on the right side, there are 8 hydrogen atoms (in H2SO4 and H2O). To balance the hydrogen, we need to add a coefficient of 4 in front of H2SO4.
k2cr2o7 + 4h2so4 + 2ki --> k2so4 + cr2(so4)3 + (10/7)h2o + I2
Now the equation becomes:
K2Cr2O7 + 4H2SO4 + 2KI --> K2SO4 + Cr2(SO4)3 + (10/7)H2O + I2
Lastly, there are 1 sulfur atom on the left side and 12 sulfur atoms on the right side (in K2SO4 and Cr2(SO4)3). To balance the sulfur, we need to add a coefficient of 8 in front of K2SO4 and a coefficient of 6 in front of Cr2(SO4)3.
k2cr2o7 + 4h2so4 + 2ki --> 8k2so4 + 6cr2(so4)3 + (10/7)h2o + I2
Now the equation is balanced:
K2Cr2O7 + 4H2SO4 + 2KI --> 8K2SO4 + 6Cr2(SO4)3 + (10/7)H2O + I2