How would you balance $H_{3} {BO}_{3}$ $"H"_3"BO"_3"$ $\to$ $rarr$ $H_{4} B_{6} O_{11} {+H}_{2} O$ $"H"_4"B"_6"O"_11" +H"_2"O"$ ?

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How would you balance $H_{3} {BO}_{3}$ $"H"_3"BO"_3"$ $\to$ $rarr$ $H_{4} B_{6} O_{11} {+H}_{2} O$ $"H"_4"B"_6"O"_11" +H"_2"O"$ ?

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Answer 1 · 2018-07-31T19:19:11

$6 l H_{3} B O_{3}$ $color(darkgreen)(6) color(white)(l)"H"_3color(black)("B")"O"_3$ $\to$ $rarr$ $1 l H_{4} B_{6} O_{11} + 7 l H_{2} O$ $color(darkgreen)(1) color(white)(l) "H"_4color(black)("B"_6) "O"_11 + color(darkgreen)(7) color(white)(l) "H"_2"O"$

Explanation:

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

$H_{3} {BO}_{3}$ $"H"_3"BO"_3$ and $H_{4} B_{6} O_{11}$ $"H"_4"B"_6"O"_11$ are the only two boron-containing species in this reaction. Each molecule of $H_{3} B O_{3}$ $"H"_3 color(navy)("B")"O"_3$ contains one boron $B$ $"B"$ atom whereas each molecule of $H_{4} B_{6} O_{11}$ $"H"_4color(navy)("B"_6)"O"_11$ contains six. Add the coefficient $6$ $"color(navy)(6)$ to the front of $H_{3} {BO}_{3}$ $"H"_3 "BO"_3$ and $1$ $color(navy)(1)$ to $H_{4} B_{6} O_{11}$ $"H"_4"B"_6"O"_11$ balance the number of boron atoms on the two sides.

$6 l H_{3} B O_{3}$ $color(purple)(6) color(white)(l)"H"_3color(navy)("B")"O"_3$ $\to$ $rarr$ $1 l H_{4} B_{6} O_{11} + H_{2} O$ $color(purple)(1) color(white)(l) "H"_4color(navy)("B"_6) "O"_11 + "H"_2"O"$ $NOT BALANCED$ $color(grey)("NOT BALANCED")$

The left-hand side of the equation contains $6 \times 3 = 18$ $6 xx 3 =18$ oxygen $O$ $"O"$ atoms. The number of oxygen atoms shall also conserve in this equation. Therefore $H_{4} B_{6} O_{11}$ $"H"_4"B"_6"O"_11$ and $H_{2} O$ $"H"_2"O"$ on the product side shall contain a total of $18$ $18$ oxygen atoms. $11$ $11$ of them go to the $H_{4} B_{6} O_{11}$ $"H"_4color(navy)("B"_6)"O"_11$ molecule. Water molecules would account for rest $7$ $7$ oxygen atoms.

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

$6 l H_{3} B O_{3}$ $color(black)(6) color(white)(l)"H"_3color(black)("B")"O"_3$ $\to$ $rarr$ $1 l H_{4} B_{6} O_{11} + 7 l H_{2} O$ $color(black)(1) color(white)(l) "H"_4color(black)("B"_6) "O"_11 + color(purple)(7) color(white)(l) "H"_2"O"$

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

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

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

$6 l H_{3} B O_{3}$ $color(darkgreen)(6) color(white)(l)"H"_3color(black)("B")"O"_3$ $\to$ $rarr$ $1 l H_{4} B_{6} O_{11} + 7 l H_{2} O$ $color(darkgreen)(1) color(white)(l) "H"_4color(black)("B"_6) "O"_11 + color(darkgreen)(7) color(white)(l) "H"_2"O"$ $Balanced$ $color(grey)("Balanced")$

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

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How would you balance $H_{3} {BO}_{3}$ $"H"_3"BO"_3"$ $\to$ $rarr$ $H_{4} B_{6} O_{11} {+H}_{2} O$ $"H"_4"B"_6"O"_11" +H"_2"O"$ ?

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How would you balance H3BO3"H"_3"BO"_3"→rarrH4B6O11 +H2O"H"_4"B"_6"O"_11" +H"_2"O"?

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How would you balance $H_{3} {BO}_{3}$ $"H"_3"BO"_3"$ $\to$ $rarr$ $H_{4} B_{6} O_{11} {+H}_{2} O$ $"H"_4"B"_6"O"_11" +H"_2"O"$ ?