fill in the gaps to balance the equation. use the smallest set of whole numbers to balance the equation and…

Explanation:

Step1: Balance Sr atoms

The number of Sr atoms on the left - hand side (LHS) in $Sr(HCO_3)_2$ is 1, and on the right - hand side (RHS) in $SrCO_3$ is 1. So the coefficient of $Sr(HCO_3)_2$ and $SrCO_3$ is 1 for Sr to be balanced. $1Sr(HCO_3)_2\rightarrow1SrCO_3 + H_2O+CO_2$

Step2: Balance H atoms

In $Sr(HCO_3)_2$, the number of H atoms is 2. In $H_2O$, to balance the H atoms, the coefficient of $H_2O$ should be 1 since there are 2 H atoms in $H_2O$. $1Sr(HCO_3)_2\rightarrow1SrCO_3 + 1H_2O+CO_2$

Step3: Balance C and O atoms

In $Sr(HCO_3)_2$, there are 2 C atoms. On the RHS, in $SrCO_3$ there is 1 C atom and in $CO_2$ there is 1 C atom. Also, for O atoms, in $Sr(HCO_3)_2$ there are 6 O atoms ($2\times3$), in $SrCO_3$ there are 3 O atoms, in $H_2O$ there is 1 O atom and in $CO_2$ there are 2 O atoms. So the balanced equation is $1Sr(HCO_3)_2\rightarrow1SrCO_3 + 1H_2O+1CO_2$

Now, for the atom - counting table:

• Sr: In the reactant $Sr(HCO_3)_2$, the number of Sr atoms is 1. In the product $SrCO_3$, the number of Sr atoms is 1.
• H: In the reactant $Sr(HCO_3)_2$, the number of H atoms is 2. In the product $H_2O$, the number of H atoms is 2.
• C: In the reactant $Sr(HCO_3)_2$, the number of C atoms is 2. In the products, 1 C atom in $SrCO_3$ and 1 C atom in $CO_2$, total 2 C atoms.
• O: In the reactant $Sr(HCO_3)_2$, the number of O atoms is 6. In the products, 3 O atoms in $SrCO_3$, 1 O atom in $H_2O$ and 2 O atoms in $CO_2$, total 6 O atoms.

The filled - in equation is: $1Sr(HCO_3)_2\rightarrow1SrCO_3 + 1H_2O+1CO_2$ The filled - in table:

Element	Reactant	Products
Sr	1	1
H	2	2
C	2	2
O	6	6

Answer:

For the equation: 1, 1, 1, 1 For the table:

Element	Reactant	Products
Sr	1	1
H	2	2
C	2	2
O	6	6