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
$"Zn" + "H"_2"SO"_4 →"ZnSO"_4 + "H"_2"SO"_4$
