how to balance this partial equation given below
k2cr2o7+h2so4+kI-->k2so4+cr2(so4)3+h2o+I2
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I will get you started.
Cr goes from +6 for each on the left to +3 on the right for each. I on the let is -1 and on the right is zero. Make things easy by making that 2KI on the left. You may need to change it later but if you don't start with that the equation never will balance.
To balance the partial equation:
1. Count the number of atoms for each element on both sides of the equation.
K: 2 on the left, 2 on the right
Cr: 2 on the left, 2 on the right
O: 7 on the left, 7 on the right
H: 2 on the left, 2 on the right
S: 1 on the left, 1 on the right
I: 1 on the left, 1 on the right
2. Start by balancing the atoms that appear in the fewest number of compounds.
Balance the K atoms by putting a coefficient of 2 in front of KI on the right:
K2Cr2O7 + H2SO4 + 2KI --> K2SO4 + Cr2(SO4)3 + H2O + I2
3. Next, the Cr atoms are balanced, so move on to the O atoms.
Balance the O atoms by putting a coefficient of 7/2 in front of H2O on the right:
K2Cr2O7 + H2SO4 + 2KI --> K2SO4 + Cr2(SO4)3 + 7/2H2O + I2
Multiply all the coefficients by 2 to get whole numbers:
2K2Cr2O7 + 2H2SO4 + 4KI --> 2K2SO4 + 2Cr2(SO4)3 + 7H2O + I2
4. Finally, check that all the atoms are balanced:
K: 4 on the left, 4 on the right
Cr: 4 on the left, 4 on the right
O: 14 on the left, 14 on the right
H: 14 on the left, 14 on the right
S: 3 on the left, 3 on the right
I: 2 on the left, 2 on the right
The balanced equation is:
2K2Cr2O7 + 2H2SO4 + 4KI --> 2K2SO4 + 2Cr2(SO4)3 + 7H2O + I2
To balance the given partial equation:
k2cr2o7 + h2so4 + kI --> k2so4 + cr2(so4)3 + h2o + I2
Follow these steps:
1. Count the number of atoms of each element on both sides of the equation.
On the left side:
- 2 atoms of K, 2 atoms of Cr, 7 atoms of O, 1 atom of S, 4 atoms of H, and 1 atom of I.
On the right side:
- 2 atoms of K, 1 atom of Cr, 4 atoms of O, 1 atom of S, 4 atoms of H, and 2 atoms of I.
2. Balance the equation by adjusting coefficients in front of the compounds. Start with the most complex or compound with the most atoms of different elements. In this case, begin with Cr2(SO4)3.
- Cr2(SO4)3 has 2 atoms of Cr, 12 atoms of O (4 x 3), 6 atoms of S (2 x 3).
- To balance Cr, place a coefficient of 2 in front of K2Cr2O7 to make it 2K2Cr2O7.
2K2Cr2O7 + h2so4 + kI --> k2so4 + cr2(so4)3 + h2o + I2
- Now the equation becomes:
2K2Cr2O7 + h2so4 + kI --> k2so4 + 2Cr2(SO4)3 + h2o + I2
3. Next, balance the sulfur (S) atoms by placing a coefficient of 3 in front of H2SO4.
2K2Cr2O7 + 3h2so4 + kI --> k2so4 + 2Cr2(SO4)3 + h2o + I2
4. Balance the hydrogen (H) atoms by placing a coefficient of 8 in front of H2O.
2K2Cr2O7 + 3h2so4 + kI --> k2so4 + 2Cr2(SO4)3 + 4h2o + I2
5. Finally, balance the iodine (I) atoms by placing a coefficient of 6 in front of KI.
2K2Cr2O7 + 3h2so4 + 6kI --> k2so4 + 2Cr2(SO4)3 + 4h2o + 3I2
The balanced equation is:
2K2Cr2O7 + 3H2SO4 + 6KI --> K2SO4 + 2Cr2(SO4)3 + 4H2O + 3I2