# How would you balance "H"_3"BO"_3"rarr"H"_4"B"_6"O"_11" +H"_2"O"?

Jul 31, 2018

$\textcolor{\mathrm{da} r k g r e e n}{6} \textcolor{w h i t e}{l} {\text{H"_3color(black)("B")"O}}_{3}$ $\rightarrow$ $\textcolor{\mathrm{da} r k g r e e n}{1} \textcolor{w h i t e}{l} \text{H"_4color(black)("B"_6) "O"_11 + color(darkgreen)(7) color(white)(l) "H"_2"O}$

#### Explanation:

This answer balances the equation by equating (the conservation of) the number of boron atoms on both sides of the equation.

${\text{H"_3"BO}}_{3}$ and ${\text{H"_4"B"_6"O}}_{11}$ are the only two boron-containing species in this reaction. Each molecule of ${\text{H"_3 color(navy)("B")"O}}_{3}$ contains one boron $\text{B}$ atom whereas each molecule of ${\text{H"_4color(navy)("B"_6)"O}}_{11}$ contains six. Add the coefficient "color(navy)(6) to the front of ${\text{H"_3 "BO}}_{3}$ and $\textcolor{n a v y}{1}$ to ${\text{H"_4"B"_6"O}}_{11}$ balance the number of boron atoms on the two sides.

$\textcolor{p u r p \le}{6} \textcolor{w h i t e}{l} {\text{H"_3color(navy)("B")"O}}_{3}$ $\rightarrow$ $\textcolor{p u r p \le}{1} \textcolor{w h i t e}{l} \text{H"_4color(navy)("B"_6) "O"_11 + "H"_2"O}$ $\textcolor{g r e y}{\text{NOT BALANCED}}$

The left-hand side of the equation contains $6 \times 3 = 18$ oxygen $\text{O}$ atoms. The number of oxygen atoms shall also conserve in this equation. Therefore ${\text{H"_4"B"_6"O}}_{11}$ and $\text{H"_2"O}$ on the product side shall contain a total of $18$ oxygen atoms. $11$ of them go to the ${\text{H"_4color(navy)("B"_6)"O}}_{11}$ molecule. Water molecules would account for rest $7$ oxygen atoms.

Each water molecule contains one single oxygen atom. $7$ of the oxygen atoms would thus correspond to $7$ water molecules on the product side. Hence the equation:

$\textcolor{b l a c k}{6} \textcolor{w h i t e}{l} {\text{H"_3color(black)("B")"O}}_{3}$ $\rightarrow$ $\textcolor{b l a c k}{1} \textcolor{w h i t e}{l} \text{H"_4color(black)("B"_6) "O"_11 + color(purple)(7) color(white)(l) "H"_2"O}$

Optionally, check if the number of hydrogen $\text{H}$ atoms on the two sides conserves to see if the equation is properly balanced:

• Number of $\text{H}$ atoms on the right-hand side: $6 \times 3 = 18$
• Number of $\text{H}$ atoms on the left-hand side: $1 \times 4 + 7 \times 2 = 18$

The two numbers are equal, and thus this chemical equation is stoichiometrically balanced.

$\textcolor{\mathrm{da} r k g r e e n}{6} \textcolor{w h i t e}{l} {\text{H"_3color(black)("B")"O}}_{3}$ $\rightarrow$ $\textcolor{\mathrm{da} r k g r e e n}{1} \textcolor{w h i t e}{l} \text{H"_4color(black)("B"_6) "O"_11 + color(darkgreen)(7) color(white)(l) "H"_2"O}$ $\textcolor{g r e y}{\text{Balanced}}$

Reference
"Balance Chemical Equation - Online Balancer," webqc.org,
https://www.webqc.org/balance.php, "Enter H3BO3 = H4B6O11 + H2O for this equation."