Write the reaction equation: Ba + BAO + Ba (OH) 2 + BaCL2

1) Ba + O2 —> BaO.

Let’s arrange the coefficients.

On the left side there are 2 oxygen atoms O, and on the right – 1 atom. Therefore, we put a coefficient of 2 in front of BaO:

Ba + O2 —> 2BaO.

Let’s also equalize the number of barium atoms. On the left there are 1, and on the right 2. Therefore, on the left in front of Ba we set the coefficient 2:

2Ba + O2 —> 2BaO.

This is the final equation.

2) ВаО + Н2О —> Ba (OH) 2.

Here the coefficients do not need to be placed, this is the final equation.

3) Ba (OH) 2 + HCl —> BaCl2 + H2O.

Let’s arrange the coefficients.

On the left side there is 1 chlorine atom Cl, and on the right side there are 2 atoms. Therefore, we put a factor of 2 in front of HCl:

Ва (OH) 2 + 2HCl —> ВаCl2 + H2O.

On the left side there are 4 hydrogen atoms H, and on the right side there are 2 atoms. Therefore, in front of H2O, we put a coefficient of 2:

Ва (OH) 2 + 2HCl —> ВаCl2 + 2H2O.

This is the final equation.

Answer: 1) 2Ba + O2 —> 2BaO; 2) BaO + H2O —> Ba (OH) 2; 3) Ba (OH) 2 + 2HCl —> BaCl2 + 2H2O.