# How do you balance Fe+H_2SO_4 -> Fe_2(SO_4)_3 + H_2?

Oct 2, 2016

$2 F {e}_{\left(s\right)} + 3 {H}_{2} S {O}_{4 \left(a q\right)} \to F {e}_{2} {\left(S {O}_{4}\right)}_{3 \left(a q\right)} +$ $3 {H}_{2}$(g)
Start by looking at the most complex compound, and balancing around that. In this case, its $F {e}_{2} {\left(S {O}_{4}\right)}_{3 \left(a q\right)}$.
We see that we need 2 Fe and 3 $S {O}_{4}$ on the reactants side, so we add those in. After that, there is only an imbalance of ${H}_{2}$, which is easily fixed by adding a 3 to the products side.