Question

balance the below equations by adding appropriate coefficients
a. CoBr3+CaSO4~CaBr2+Co2(SO4)3

Answer 1

To balance the equation \( \text{CoBr}_3 + \text{CaSO}_4 \rightarrow \text{CaBr}_2 + \text{Co}_2(\text{SO}_4)_3 \), we need to ensure that the number of atoms of each element on the reactant side equals the number on the product side.

Let's write the unbalanced equation:

\[ \text{CoBr}_3 + \text{CaSO}_4 \rightarrow \text{CaBr}_2 + \text{Co}_2(\text{SO}_4)_3 \]

1. Count the number of each atom on both sides:
- Reactants:
- Co: 1
- Br: 3
- Ca: 1
- S: 1
- O: 4
- Products:
- Co: 2
- Br: 2
- Ca: 1
- S: 3
- O: 12 (from \( \text{SO}_4 \))

2. Next, let’s adjust the coefficients to balance the equation.

Start with Co. We have 2 Co on the product side, so we need 2 CoBr\(_3\):

\[ 2 \text{CoBr}_3 + \text{CaSO}_4 \rightarrow \text{CaBr}_2 + \text{Co}_2(\text{SO}_4)_3 \]

Now let’s recount:
- Co: 2 (reactants) 2 (products)
- Br: 6 (reactants, \( 2 \times 3 \)) and 2 (products)
- Ca: 1 for both sides
- S: 1 (\( \text{CaSO}_4 \)) and 3 (\( \text{Co}_2(\text{SO}_4)_3 \))
- O: 4 (reactants) and 12 (products)

To balance Br, we add a coefficient of 3 in front of CaBr\(_2\):

Now the equation looks like:
\[ 2 \text{CoBr}_3 + \text{CaSO}_4 \rightarrow 3 \text{CaBr}_2 + \text{Co}_2(\text{SO}_4)_3 \]

Recounting gives us:
1. Co: 2 (reactants) 2 (products) - balanced.
2. Br: 6 (reactants) and 6 (products, \( 3 \times 2 = 6 \)) - balanced.
3. Ca: 1 and 3 - not balanced.

To balance Ca, we’ll need 3 \( \text{CaSO}_4 \) on the reactant side.
So we have:
\[ 2 \text{CoBr}_3 + 3 \text{CaSO}_4 \rightarrow 3 \text{CaBr}_2 + \text{Co}_2(\text{SO}_4)_3 \]

Final count:
- Co: 2 in products, 2 in reactants.
- Br: 6 in reactants, 6 in products.
- Ca: 3 in reactants and products.
- S: 3 and 3 in both sides
- O: 12 and 12 in both sides.

The balanced equation is:

\[ 2 \text{CoBr}_3 + 3 \text{CaSO}_4 \rightarrow 3 \text{CaBr}_2 + \text{Co}_2(\text{SO}_4)_3 \]