Question

S+2H2SO4=3S+O2+2H2O ovr

Answer 1

Ответ:S + 2H2SO4 = 3SO2 + 2H2O

Объяснение: S(0) -4e = S(+4)

S(+6) + 2e = S(+4)     НОК: 4, следовательно коэффициент перед Серой в степени окисления (0) - единица, а перед серой в степени окисления +6 - двойка.

Уравниваем количество серы с правой стороны (перед SO4) коэффициент 3 , и уравниваем количество атомов оксигена и водорода (двойка перед водой) .

Answer 2

The chemical equation you provided appears to be unbalanced. To balance it, you need to ensure that the same number of each type of atom is on both sides of the equation. Let's balance the equation for the reaction:

S + 2H2SO4 → 3S + O2 + 2H2O

First, let's balance the sulfur (S) atoms:

On the left side, there's 1 sulfur atom, and on the right side, there are 3 sulfur atoms. To balance sulfur, you can put a coefficient of 3 in front of S on the left side:

3S + 2H2SO4 → 3S + O2 + 2H2O

Now, sulfur is balanced, but we need to balance the hydrogen (H) and oxygen (O) atoms. There are 4 hydrogen atoms on the left side and 4 hydrogen atoms on the right side, so hydrogen is already balanced. However, there are 4 oxygen atoms on the right side (2 from O2 and 2 from H2O) and only 2 oxygen atoms on the left side. To balance the oxygen atoms, you can put a coefficient of 5/2 (which can be written as 2.5) in front of O2 on the left side:

3S + 2H2SO4 → 3S + 2.5O2 + 2H2O

Now, the equation is balanced in terms of the number of each type of atom. However, it's not common to have fractional coefficients in a balanced chemical equation. To eliminate the fractional coefficient, you can multiply the entire equation by 2 to get whole number coefficients:

2(3S + 2H2SO4) → 2(3S + 2.5O2 + 2H2O)

This results in:

6S + 4H2SO4 → 6S + 5O2 + 4H2O

Now, the equation is balanced with whole number coefficients:

6S + 4H2SO4 → 6S + 5O2 + 4H2O