How would I balance this equation Ca(C2H3O2)2+K3PO4-->Ca3(PO4)2+KC2H3O2
My advice is to pick what appears to be the most complicated formula, and start there. Ca3(PO4)2 appears to be it.
So you need three Ca, so put a 3 on the left side in front of the calcium acetate.
YOu need two phosphates, so put a 2 in front on the left of the potassium phosphate.
Now, all that is left is to put the right coefficant on the right before postassium acetate.
3Ca(C2H3O2)+2K3PO4=1Ca3(PO4)2+6KC2H3O2
To balance the equation:
Ca(C2H3O2)2 + K3PO4 → Ca3(PO4)2 + KC2H3O2
1. Start by counting the number of atoms on each side of the equation.
On the left side:
Calcium (Ca): 1
Carbon (C): 2 (from C2H3O2)
Hydrogen (H): 6 (from C2H3O2)
Oxygen (O): 4 (from C2H3O2)
On the right side:
Calcium (Ca): 3 (from Ca3(PO4)2)
Phosphorus (P): 2 (from Ca3(PO4)2)
Oxygen (O): 8 (from Ca3(PO4)2)
Potassium (K) and Acetate (C2H3O2) are balanced already.
2. Balance the phosphorus (P) atoms by placing a coefficient of 3 in front of K3PO4 and a coefficient of 2 in front of Ca3(PO4)2:
Ca(C2H3O2)2 + 3K3PO4 → 2Ca3(PO4)2 + KC2H3O2
Now, the phosphorus (P) atoms are balanced.
3. Balance the calcium (Ca) atoms by placing a coefficient of 3 in front of Ca(C2H3O2)2:
3Ca(C2H3O2)2 + 3K3PO4 → 2Ca3(PO4)2 + KC2H3O2
Now, the calcium (Ca) atoms are balanced.
4. Balance the hydrogen (H) atoms by placing a coefficient of 6 in front of KC2H3O2:
3Ca(C2H3O2)2 + 3K3PO4 → 2Ca3(PO4)2 + 6KC2H3O2
Now, the hydrogen (H) atoms are balanced.
5. Finally, balance the oxygen (O) atoms by adjusting the coefficients:
3Ca(C2H3O2)2 + 3K3PO4 → 2Ca3(PO4)2 + 6KC2H3O2
There are now:
Calcium (Ca): 3
Carbon (C): 6
Hydrogen (H): 12
Oxygen (O): 24
Potassium (K): 18
Phosphorus (P): 6
And the equation is balanced.
To balance the chemical equation Ca(C2H3O2)2 + K3PO4 --> Ca3(PO4)2 + KC2H3O2, we need to make sure that there are an equal number of atoms on both sides of the equation. Here's how you can balance it:
1. Start by counting the number of atoms on each side of the equation.
On the left side, there are 1 Calcium (Ca), 4 Carbon (C), 4 Hydrogen (H), 4 Oxygen (O), and 1 Potassium (K).
On the right side, there are 3 Calcium (Ca), 2 Phosphorus (P), and 7 Oxygen (O).
2. Balance the atoms one by one, starting with the elements that appear in the fewest compounds. In this equation, we can start by balancing the Calcium (Ca) atoms.
Counting the number of Calcium (Ca) atoms, we have 1 on the left side but 3 on the right side. To balance the Ca atoms, we place a 3 in front of Ca(C2H3O2)2 on the reactant side and a 1 in front of Ca3(PO4)2 on the product side:
3Ca(C2H3O2)2 + K3PO4 --> Ca3(PO4)2 + KC2H3O2
3. Counting the Carbon (C) and Hydrogen (H) atoms, we have 6 C and 6 H on the left side, so we need to balance them on the right side.
We can achieve this by placing a 2 in front of KC2H3O2 on the product side:
3Ca(C2H3O2)2 + K3PO4 --> Ca3(PO4)2 + 2KC2H3O2
4. Counting the Oxygen (O) atoms, we have 12 on the left side and 18 on the right side.
We can balance the Oxygen atoms by placing a coefficient of 6 in front of K3PO4 on the reactant side:
3Ca(C2H3O2)2 + 6K3PO4 --> Ca3(PO4)2 + 2KC2H3O2
5. Finally, counting the Phosphorus (P) atoms, we have 6 on the right side, so we need to balance them on the left side.
To balance the Phosphorus atoms, we place a coefficient of 2 in front of Ca3(PO4)2 on the product side:
3Ca(C2H3O2)2 + 6K3PO4 --> 2Ca3(PO4)2 + 2KC2H3O2
Now the equation is balanced with an equal number of atoms on both sides.