Write the balanced equation for the neutralization of HCl(aq) and [latex]Ba(OH)_{2}(s)[/latex].

STEP 1 Write the reactants and products. [latex] HCl(aq) + Ba(OH)_{2}(s) → H_{2}O(l) + salt [/latex] STEP 2 Balance the [latex]H^{+}[/latex] in the acid with the [latex] OH^{-}[/latex] in the base. Placing a coefficient of 2 in front of the HCl provides [latex]2H^{+}[/latex] for the [latex]2OH^{-}[/latex] from [latex]Ba(OH)_{2}[/latex]. [latex]2HCl(aq) + Ba(OH_{2})(s) → H_{2}O(l) + salt [/latex] STEP 3 Balance the [latex]H_{2}O[/latex] with the [latex] H^{+}[/latex] and the [latex]OH^{-}[/latex] . Use a coefficient of 2 in front of [latex]H_{2}O[/latex] to balance [latex]2H^{+}[/latex] and [latex]2OH^{-}[/latex]. [latex] 2HCl(aq) + Ba(OH)_{2}(s) → 2H_{2}O(l) + salt [/latex] STEP 4 Write the salt from the remaining ions. Use the ions [latex]Ba^{2+}[/latex] and [latex]2Cl^{-}[/latex] and write the formula for the salt as [latex]BaCl_{2}[/latex]. [latex]2HCl(aq) + Ba(OH)_{2}(s) → 2H_{2}O(l) + BaCl_{2}(aq)[/latex]

Question 11.11: Write the balanced equation for the neutralization of HCl(aq...

General, Organic, and Biological Chemistry: Structures of Life

Write the balanced equation for the neutralization of HCl(aq) and $Ba(OH)_{2}(s)$ .

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STEP 1 Write the reactants and products.

HCl(aq) + Ba(OH)_{2}(s) → H_{2}O(l) + salt

STEP 2 Balance the $H^{+}$ in the acid with the $OH^{-}$ in the base. Placing a coefficient of 2 in front of the HCl provides $2H^{+}$ for the $2OH^{-}$ from $Ba(OH)_{2}$ .
$2HCl(aq) + Ba(OH_{2})(s) → H_{2}O(l) + salt$

STEP 3 Balance the $H_{2}O$ with the $H^{+}$ and the $OH^{-}$ . Use a coefficient of 2 in front of $H_{2}O$ to balance $2H^{+}$ and $2OH^{-}$ .
$2HCl(aq) + Ba(OH)_{2}(s) → 2H_{2}O(l) + salt$

STEP 4 Write the salt from the remaining ions. Use the ions $Ba^{2+}$ and $2Cl^{-}$ and write the formula for the salt as $BaCl_{2}$ .
$2HCl(aq) + Ba(OH)_{2}(s) → 2H_{2}O(l) + BaCl_{2}(aq)$