You follow a systematic procedure to balance the equation.
Start with the unbalanced equation:
#"Zn" + "H"_2"SO"_4 → "ZnSO"_4 +"H"_2#
A method that often works is first to balance everything other than #"O"# and #"H"#, 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 #"ZnSO"_4#. We put a 1 in front of it to remind ourselves that the number is now fixed.
We start with
#"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#
Balance #"Zn"#:
We have #"1 Zn"# on the right, so we need #"1 Zn"# on the left. We put a 1 in front of the #"Zn"#.
#color(blue)(1)"Zn" + "H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#
Balance #"S"#:
We have fixed #"1 S"# on the right. We need #"1 S"# on the left. Put a 1 in front of #"H"_2"SO"_4#.
#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 +"H"_2#
Balance #"O"#:
Done.
Balance #"H"#:
We have fixed #"2 H"# on the left, so we need #"2 H"# on the right. Put a 1 in front of #"H"_2#.
#color(blue)(1)"Zn" + color(orange)(1)"H"_2"SO"_4 → color(red)(1)"ZnSO"_4 + color(purple)(1)"H"_2#
Every formula now has a coefficient. We should have a balanced equation.
Let's check.
#bb("Atom" color(white)(m)"lhs"color(white)(m)"rhs")#
#color(white)(m)"Zn"color(white)(mml)1color(white)(mml)1#
#color(white)(m)"H"color(white)(mmll)2color(white)(mml)2#
#color(white)(m)"S"color(white)(mmm)1color(white)(mml)1#
#color(white)(m)"O"color(white)(mmll)4color(white)(mml)4#
All atoms balance. The balanced equation is
#"Zn" + "H"_2"SO"_4 →"ZnSO"_4 + "H"_2"SO"_4#
