complete and balance the following equation:
CoCl*6H2O+ NH4Cl+ NH3+ H2O --->
[Co(NH3)5Cl]Cl2+ H2O
To complete and balance the equation, we need to ensure that the number of atoms on both sides of the chemical equation is the same, and that the charges are also balanced. Let's go step by step:
1. Write down the unbalanced equation:
CoCl*6H2O + NH4Cl + NH3 + H2O → [Co(NH3)5Cl]Cl2 + H2O
2. Begin balancing the equation by looking at the coefficients in front of each compound. In this case, we only have CoCl*6H2O as the compound with a subscript number. Let's separate the coefficient and the compound:
CoCl * 6H2O + NH4Cl + NH3 + H2O → [Co(NH3)5Cl]Cl2 + H2O
3. Next, let's balance the Cobalt (Co) atoms. We have 1 Co atom on the left side, so we need 1 Co atom on the right side:
CoCl * 6H2O + NH4Cl + NH3 + H2O → [Co(NH3)5Cl]Cl2 + H2O
4. Now, let's balance the Chlorine (Cl) atoms. On the left side, we have 1 Cl atom in CoCl*6H2O and 1 Cl atom in NH4Cl. So, we need 2 Cl atoms on the right side:
CoCl * 6H2O + NH4Cl + NH3 + H2O → [Co(NH3)5Cl]Cl2 + H2O
5. Balance the Hydrogen (H) atoms. On the left side, we have 12 H atoms in 6H2O and 4 H atoms in NH4Cl. So, we need 16 H atoms on the right side:
CoCl * 6H2O + NH4Cl + NH3 + H2O → [Co(NH3)5Cl]Cl2 + 16 H2O
6. Next, balance the Nitrogen (N) atoms. On the left side, we have 1 N atom in NH4Cl and 1 N atom in NH3. So, we need 2 N atoms on the right side:
CoCl * 6H2O + NH4Cl + NH3 + H2O → [Co(NH3)5Cl]Cl2 + 16 H2O
7. Finally, balance the Oxygen (O) atoms. On the left side, we have 6 O atoms in 6H2O. On the right side, we have 16 O atoms in 16 H2O. So, the O atoms are already balanced.
The balanced equation is:
CoCl * 6H2O + NH4Cl + NH3 + H2O → [Co(NH3)5Cl]Cl2 + 16 H2O
To complete and balance the equation, we need to ensure that the number of atoms on both sides of the equation is equal. Let's balance the equation step by step.
The unbalanced equation is:
CoCl*6H2O + NH4Cl + NH3 + H2O ---> [Co(NH3)5Cl]Cl2 + H2O
Step 1: Balance the number of metal atoms
Since there is only one cobalt (Co) atom on the left side, we need to have one cobalt atom on the right side. We can achieve this by adding a coefficient of 2 in front of CoCl*6H2O:
2CoCl*6H2O + NH4Cl + NH3 + H2O ---> [Co(NH3)5Cl]Cl2 + H2O
Step 2: Balance the number of chloride ions (Cl-)
On the left side, we have 6 chloride ions from CoCl*6H2O, 1 chloride ion from NH4Cl, and none from NH3. On the right side, we have 2 chloride ions from [Co(NH3)5Cl]Cl2. To balance the chloride ions, we need to add a coefficient of 3 in front of NH4Cl:
2CoCl*6H2O + 3NH4Cl + NH3 + H2O ---> [Co(NH3)5Cl]Cl2 + H2O
Step 3: Balance the number of ammonia molecules (NH3)
On the left side, we have 1 NH3 molecule, and on the right side, we have 1 NH3 molecule in [Co(NH3)5Cl]Cl2. So, the number of NH3 molecules is already balanced.
2CoCl*6H2O + 3NH4Cl + NH3 + H2O ---> [Co(NH3)5Cl]Cl2 + H2O
Step 4: Balance the number of water molecules (H2O)
On the left side, we have 6 H2O molecules from CoCl*6H2O and 1 H2O molecule separately. On the right side, we have 1 H2O molecule in [Co(NH3)5Cl]Cl2. To balance the water molecules, we need to add a coefficient of 5 in front of H2O:
2CoCl*6H2O + 3NH4Cl + NH3 + 5H2O ---> [Co(NH3)5Cl]Cl2 + H2O
Finally, the balanced equation is:
2CoCl*6H2O + 3NH4Cl + NH3 + 5H2O ---> [Co(NH3)5Cl]Cl2 + H2O