How would you balance this chemical equation: #2Sb + I_2 -> 2SbI_3#?

There are six Iodines in the products so there must be six iodines in the reactants.

Iodine comes as diatomic gas molecule. ( as well as a grey crystal).

# 3 xx I_2 = 6 I^-1# therefore it requires 3 #I_2# molecules to balance the equation

Start with the unbalanced equation:

#"2Sb" + "I"_2 → "2SbI"_3#

A method that often works is to balance everything other than #"O"# and #"H"# first, then balance #"O"#, and finally balance #"H"#.

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like #"SbI"_3#. It already has a 2 in front of it.

#"2Sb" + "I"_2 → color(red)(2)"SbI"_3#

Balance #"Sb"#:

There are #"2 Sb"# on the right, and there are #"2 Sb"# on the right. #"Sb"# is balanced.

#color(blue)(2)"Sb" + "I"_2 → color(red)(2)"SbI"_3#

Balance #"I"#:

We have fixed #"6 I"# on the right, so we need #"6 I"# on the left. Put a 3 in front of #"I"_2#.

#color(blue)(2)"Sb" + color(orange)(3)"I"_2 → color(red)(2)"SbI"_3#

Every formula now has a fixed coefficient. We should have a balanced equation.

Let’s check:

#bb("Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side")#
#color(white)(m)"Sb"color(white)(mmmmm)2color(white)(mmmmmmmm)2#
#color(white)(m)"I"color(white)(mmmmmm)6color(white)(mmmmmmmm)6#

All atoms balance. The balanced equation is

#2"Sb" + 3"I"_2 → 2"SbI"_3#