How do you balance #Mn + HI -> H_2 + MnI_3#?

The procedure I usually use is:

1. Start with the most complicated formula.
2. Balance all atoms other than #"H"# and #"O"#.
3. Balance #"O"#.
4. Balance #"H"#.
5. If necessary, multiply by an integer to remove fractions.
6. Check that all atoms are balanced.

Your unbalanced equation is

#"Mn"color(white)(l) + "HI" → "H"_2 + "MnI"_3#

Step 1. Start with the most complicated formula.

Put a 1 in front of #"MnI"_3#. This number is fixed and does not change until (if necessary) Step 5.

#"Mn"color(white)(l) + "HI" → "H"_2 + color(red)(1)"MnI"_3#

Step 2. Balance #"Mn"#.

We have fixed #"1 Mn"# on the right, so we need #"1 Mn"# on the left.

#color(blue)(1)"Mn"+ "HI" → "H"_2 + color(red)(1)"MnI"_3#

Step 3. Balance #"I"#.

We have fixed #"3 I"# on the right, so we need #"3 I"# on the left.

Put a 3 before #"HI"#.

#color(blue)(1)"Mn"+ color(orange)(3)"HI" → "H"_2 + color(red)(1)"MnI"_3#

Step 4. Balance #"H"#.

We have fixed #"3 H"# on the left, so we need #"3 H"# on the right.

We have #"2 H"# atoms on the right, so we must multiply by 1.5 to get #"3 H"# atoms.

#color(blue)(1)"Mn"+ color(orange)(3)"HI" → color(green)(1.5)"H"_2 + color(red)(1)"MnI"_3#

All atoms should now between balanced, but we have a fraction in front of #"H"_2#

Step 5. Multiply all coefficients by 2 to remove fractions.

#2"Mn"+ 6"HI" → 3"H"_2 + "MnI"_3#

Step 6. Check that all atoms are balanced.

#bb"On the left"color(white)(l) bb"On the right"#
#color(white)(mll)"2 Mn"color(white)(mmmm) "2 Mn"#
#color(white)(mll)"6 H"color(white)(mmmmm) "6 H"#
#color(white)(mll)"6 I"color(white)(mmmmml) "6 I"#

The balanced equation is

#color(red)(2"Mn"+ 6"HI" → 3"H"_2 + "MnI"_3)#