hydrogen sulfide reacts with sulphur dioxide to give water and S, balance the equation. if hydrogen sulfide contains 125g, how much S is produced?

2H2S+SO2>>> 2H2O+2S

for each 2 moles of H2S, two moles of S are produced.

moles of H2S=125/34
so, you should get the same number of moles of S, or grams S=125/34 * 32

To balance the chemical equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation.

The balanced equation for the reaction between hydrogen sulfide (H2S) and sulfur dioxide (SO2) to give water (H2O) and sulfur (S) is as follows:

2H2S + SO2 → 3H2O + 3S

Now, let's calculate the amount of sulfur (S) produced when 125g of hydrogen sulfide (H2S) reacts.

We can start by converting the given mass of hydrogen sulfide (H2S) to the number of moles using its molar mass. The molar mass of hydrogen sulfide (H2S) = 2(atomic mass of hydrogen) + atomic mass of sulfur.

Molar mass of hydrogen (H) = 1 g/mol
Molar mass of sulfur (S) = 32 g/mol

Molar mass of H2S = 2(1) + 32 = 34 g/mol

Number of moles of H2S = Mass of H2S / Molar mass of H2S
= 125 g / 34 g/mol
≈ 3.676 mol

According to the balanced equation, 2 moles of hydrogen sulfide react to produce 3 moles of sulfur.

Therefore, we can set up a proportion:

2 moles H2S / 3 moles S = 3.676 moles H2S / x moles S

Cross-multiplying and solving for x:

2x = 3 * 3.676
2x = 11.028
x = 11.028 / 2
x ≈ 5.514

So, approximately 5.514 moles of sulfur (S) are produced when 125g of hydrogen sulfide (H2S) reacts.

To calculate the mass of sulfur produced, we can use the molar mass of sulfur (32 g/mol):

Mass of S = Number of moles of S × Molar mass of S
= 5.514 mol × 32 g/mol
≈ 176.448 g

Therefore, approximately 176.448 g of sulfur (S) is produced when 125g of hydrogen sulfide (H2S) reacts.