Aqueous solutions of copper (II) sulfate, CuSO4, and potassium carbonate, K2CO3, a blue parcipitate of copper (II) carbonate forms in addition to one other compound.
We have to write the balanced chemical equation for that but our teacher didn't teach us anything about that
CuSO4(aq) + K2CO3(aq) ==> CuCO3(s) + K2SO4(aq)
To write a balanced chemical equation, we need to identify the reactants and products involved in the reaction.
In this case, the reactants are copper(II) sulfate (CuSO4) and potassium carbonate (K2CO3). The products are copper(II) carbonate (CuCO3) and potassium sulfate (K2SO4).
The first step is to write the unbalanced equation:
CuSO4 + K2CO3 → CuCO3 + K2SO4
Next, we need to balance the equation by ensuring that the number of atoms of each element is the same on both sides of the equation.
First, let's balance the copper (Cu) atoms:
CuSO4 + K2CO3 → CuCO3 + K2SO4
Next, let's balance the sulfur (S) atoms:
CuSO4 + K2CO3 → CuCO3 + K2SO4
Now, let's balance the oxygen (O) atoms:
CuSO4 + K2CO3 → CuCO3 + K2SO4
Since all the atoms are balanced, the final balanced chemical equation is:
CuSO4 + K2CO3 → CuCO3 + K2SO4
No problem! I can explain how to write a balanced chemical equation for this reaction.
To start, let's first identify the reactants and products in the reaction:
Reactants:
1. Copper (II) sulfate (CuSO4)
2. Potassium carbonate (K2CO3)
Products:
1. Copper (II) carbonate (CuCO3)
2. A compound (we don't know its chemical formula yet)
Now, let's proceed to write the balanced chemical equation step by step:
Step 1: Write the chemical formulas of the reactants and products as the unbalanced equation:
CuSO4 + K2CO3 → CuCO3 + ?
Step 2: Determine the number of atoms of each element on both sides of the equation and balance them. Start with elements that appear in only one compound on each side of the equation.
Element: Copper (Cu)
On the left side: 1 Cu
On the right side: 1 Cu
The number of Cu atoms is already balanced.
Element: Sulfur (S)
On the left side: 1 S
On the right side: 1 S
The number of S atoms is already balanced.
Element: Oxygen (O)
On the left side: 4 O (2 from CuSO4 and 2 from K2CO3)
On the right side: 3 O (from CuCO3)
The number of O atoms is not balanced. To balance it, we need to add a coefficient in front of CuCO3. Let's put a "2" in front of CuCO3:
CuSO4 + K2CO3 → CuCO3 + ?
CuSO4 + K2CO3 → 2CuCO3 + ?
Element: Potassium (K)
On the left side: 2 K (from K2CO3)
On the right side: 0 K
The number of K atoms is not balanced. To balance it, we need to add a coefficient in front of K2CO3. Let's put a "2" in front of K2CO3:
CuSO4 + 2K2CO3 → 2CuCO3 + ?
Now, the number of K atoms is balanced.
Step 3: Determine the compound formed on the right side.
Since we only know that it is another compound, we can represent it as "Compound X" for now. The equation becomes:
CuSO4 + 2K2CO3 → 2CuCO3 + Compound X
And there you have it! The balanced chemical equation for the reaction of copper (II) sulfate and potassium carbonate forming copper (II) carbonate and another compound is:
CuSO4 + 2K2CO3 → 2CuCO3 + Compound X
Note: To determine the specific compound formed, you may need additional information or experimental data.