hydrogen sulfide reacts with sulphur dioxide to give water and S, balance the equation. if hydrogen sulfide contains 125g, how much S is produced?
Don't you follow up on your posts. Bob Pursley answered your first post hours ago.
To balance the chemical equation, we need to make sure that the number of atoms of each element is the same on both sides.
The balanced equation for the reaction between hydrogen sulfide (H2S) and sulfur dioxide (SO2) to produce water (H2O) and sulfur (S) is:
H2S + SO2 → H2O + S
Now, let's calculate the amount of S (sulfur) that is produced when 125g of H2S reacts.
1. Convert the mass of H2S to moles:
We need to determine the number of moles of H2S. To do this, we'll use the molar mass of H2S, which is the sum of the atomic masses of its constituents (H = 1.01 g/mol, S = 32.07 g/mol):
Molar mass of H2S = 2(1.01 g/mol) + 32.07 g/mol = 34.09 g/mol
Now, we can calculate the moles of H2S:
Moles of H2S = mass of H2S / molar mass of H2S
= 125 g / 34.09 g/mol
≈ 3.665 mol
2. Use the balanced equation to determine the ratio of moles between H2S and S:
From the balanced equation, we see that 1 mole of H2S produces 1 mole of S. Therefore, the molar ratio between H2S and S is 1:1.
3. Calculate the moles of S produced:
Since the ratio of moles between H2S and S is 1:1, the moles of S produced will also be 3.665 mol.
4. Convert moles of S to mass:
To determine the mass of S produced, we multiply the moles of S by its molar mass (S = 32.07 g/mol):
Mass of S = moles of S × molar mass of S
= 3.665 mol × 32.07 g/mol
≈ 117.34 g
Thus, approximately 117.34 grams of sulfur (S) will be produced when 125 grams of hydrogen sulfide (H2S) reacts with sulfur dioxide (SO2) to form water (H2O).