The procedure I usually use is:
- Start with the most complicated formula.
- Balance all atoms other than "H" and "O".
- Balance "O".
- Balance "H".
- If necessary, multiply by an integer to remove fractions.
- 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)