How do you balance #K+B_2O_3 -> K_2O + B#?

1 Answer
Mar 6, 2016

Answer:

One way is to use the method of oxidation numbers.

Explanation:

We start with the unbalanced equation:

#"K" + "B"_2"O"_3 → "K"_2"O" + "B"#

Step 1. Identify the atoms that change oxidation number

#stackrelcolor(blue)(0)("K") + stackrelcolor(blue)(+3)("B")_2stackrelcolor(blue)("-2")("O")_3 → stackrelcolor(blue)(+1)("K")_2stackrelcolor(blue)("-2")("O") + stackrelcolor(blue)(0)("B")#

The changes in oxidation number are:

#"K: 0 → +1; Change =+1 (oxidation)"#
#"B: +3 → 0; Change = -3 (reduction)"#

Step 2. Equalize the changes in oxidation number

We need 3 atoms of #"K"# for every 1 atom of #"B"# or 6 atoms of #"K"# for every 2 atoms of #"B"#. This gives us total changes of +6 and -6.

Step 3. Insert coefficients to get these numbers

#color(red)(6)"K" + color(red)(1)"B"_2"O"_3 → color(red)(3)"K"_2"O" + color(red)(2)"B"#

Every formula now has a coefficient. The equation should be balanced.

Step 4. Check that all atoms are balanced.

#bb"On the left"color(white)(l)bb "On the right"#
#color(white)(mm)"6 K"color(white)(mmmm) "6 K"#
#color(white)(mm)"2 B"color(white)(mmmm) "2 B"#
#color(white)(mm)"3 O"color(white)(mmmm) "3 O"#

The balanced equation is

#color(red)(6"K" + "B"_2"O"_3 → 3"K"_2"O" + 2"B"#