We can use the method of oxidation numbers to balance this equation.
We start with the unbalanced equation:
#"P"_4 + "H"_2"S""O"_4 → "H"_3"P""O"_4 + "H"_2"O" + "S""O"_2#
Step 1. Identify the atoms that change oxidation number
#stackrelcolor(blue)(0)("P")_4 + stackrelcolor(blue)("+1")"H"_2stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 → stackrelcolor(blue)("+1")("H")_3stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") + stackrelcolor(blue)("+4")("S")stackrelcolor(blue)("-2")("O")_2#
The changes in oxidation number are:
#"P:" color(white)(m)0 → "+5"; "Change = +5 (oxidation)"#
#"S: +6 → +4; Change ="color(white)(m) "-2 (reduction)"#
Step 2. Equalize the changes in oxidation number
We need 5 atoms of #"S"# for every 2 atoms of #"P"# or 10 atoms of #"S"# for every 4 atoms of #"P"#. This gives us total changes of -20 and +20.
Step 3. Insert coefficients to get these numbers
#color(red)(1)"P"_4 + color(red)(10)"H"_2"S""O"_4 → color(red)(4)"H"_3"P""O"_4 + "H"_2"O" + color(red)(10)"S""O"_2#
Step 4. Balance #"O"#
We have fixed 40 #"O"# atoms on the left and 36 #"O"# atoms on the right, so we need 4 more #"O"# atoms on the right. Put a 4 before #"H"_2"O"#.
#color(red)(1)"P"_4 + color(red)(10)"H"_2"S""O"_4 → color(red)(4)"H"_3"P""O"_4 + color(blue)(4)"H"_2"O" + color(red)(10)"S""O"_2#
Every formula now has a coefficient. The equation should be balanced.
Step 7. Check that all atoms are balanced.
#bb("On the left"color(white)(l) "On the right")#
#color(white)(mmll)"4 P"color(white)(mmmmm) "4 P"#
#color(white)(mm)"20 H"color(white)(mmmml) "20 H"#
#color(white)(mm)"10 S"color(white)(mmmmll) "10 S"#
#color(white)(mm)"40 O"color(white)(mmmml) "40 O"#
The balanced equation is
#color(red)("P"_4 + "10H"_2"S""O"_4 → "4H"_3"P""O"_4 + "4H"_2"O" + "10S""O"_2)#
