Question

mg + hcl → mgcl₂ + h₂
total # before (reactants) elements total # after (products)
mg
h
cl

Explanation:

Step1: Balance Mg

The number of Mg atoms on the left - hand side (reactants) and right - hand side (products) should be equal. There is 1 Mg atom in $MgCl_2$ on the product side, so we put 1 in front of Mg on the reactant side: $1Mg+ \underline{\hspace{1cm}}HCl
ightarrow1MgCl_2 + \underline{\hspace{1cm}}H_2$.

Step2: Balance Cl

In $MgCl_2$, there are 2 Cl atoms. Since each HCl molecule has 1 Cl atom, we need 2 HCl molecules on the reactant side to balance the Cl atoms: $1Mg + 2HCl
ightarrow1MgCl_2+\underline{\hspace{1cm}}H_2$.

Step3: Balance H

On the reactant side, we have 2 H atoms from 2 HCl. On the product side, in $H_2$, there are also 2 H atoms. So the balanced equation is $1Mg + 2HCl
ightarrow1MgCl_2+1H_2$.

Now fill in the table:

Total # Before (reactants)	Elements	Total # After (Products)
2	H	2
2	Cl	2

Answer:

The balanced equation is $1Mg + 2HCl
ightarrow1MgCl_2+1H_2$ and the table values are: for Mg, 1 before and 1 after; for H, 2 before and 2 after; for Cl, 2 before and 2 after.