Given reaction: $$Mgleft(OH\right)_{2}+HCl\rightarrow MgCl_{2}+H_{2}O$$ In the above reaction, $$Mg(OH)_2$$ and $$HCl$$ are reactants, and $$Mg Cl_2$$ and $$H_2O$$ are the products. $$ extbf{Step 1: Count the number of atoms of each element on either side of the reaction}$$  Reactant side Product side Mg = 1Mg = 1 H = 3H = 2 O = 2O = 1 Cl = 1Cl = 2 $$ extbf{Step 2: Multiply HCl in the reactant side by 2}$$ Since one extra chlorine is present on the product side. Thus, multiply HCl by 2 in order to balance the Cl-atoms. $$Mgleft(OH\right)_{2}+2HCl\rightarrow MgCl_{2}+H_{2}O$$ $$ extbf{Step 3: Balance H and O atoms on both the sides}$$ Now, two extra hydrogen and one extra oxygen atoms are present on the reactant side. Thus, multiply $$H_{2}O$$ on the product side by 2 in order to balance the H and O atoms. $$Mgleft(OH\right)_{2}+2HCl\rightarrow MgCl_{2}+2H_{2}O$$ Therefore, the balanced chemical equation is $$Mgleft(OH\right)_{2}+2HCl\rightarrow MgCl_{2}+2H_{2}O$$