Answer:
2H2S + 3O2 --> 2H2O + 2SO2
2 mol H2S/8 mol H2O = 0.25 mol
Explanation:
Reactants: H is 2; S is 1; O is 2
Products: H is 2; S is 1; O is 3
There are 3 oxygens so add a 2 to the H2O e and 3 to the reactants side.
H2S + 3O2 → 2H2O + SO2
Reactant side: H is 2; S is 1; O is 6
Product side: H is 4; S is 1; O is 4
Hydrogen is now out of balance, so a coefficient of 2 can be placed in front of H2S to give:
2H2S + 3O2 → 2H2O + SO2
Reactant side: H is 4; S is 2; O is 6
Product side: H is 4; S is 1; O is 4
Sulfur is unbalanced, so a coefficient of 2 can be placed in front of SO2 to give:
2H2S + 3O2 → 2H2O + 2SO2
